Slope field for differential equation with cube root: "Missing number, treated as zero"

Question

Since pgfplots doesn't have a "built in" cube root function, I adapted code from this question in order to plot a cube root function in a differential equation slope field. The trick seems to display perfectly:

However, although compiling and looking perfectly as desired, my Overleaf project is returning a "missing number, treated as zero" error. Here is the MWE:

\documentclass{exam}
\printanswers
\usepackage{pgf,tikz}
\usepackage{pgfplots}
\usepackage{physics}
%Slope fields, from https://tex.stackexchange.com/questions/139064/how-to-draw-slope-fields-with-all-the-possible-solution-curves-in-latex
\pgfplotsset{ODE/.style={
    axis x line = middle,
    axis y line = middle,
    xlabel={$x$}, ylabel={$y$},
    axis equal image, % Unit vectors for both axes have the same length
    view={0}{90}, % We need to use "3D" plots, but we set the view so we look at them from straight up
    samples=21, % How many arrows?
    cycle list={    % Plot styles
            blue,
            quiver={
                u={1/\length}, v={f(x)/\length}, % End points of the arrows
                    scale arrows=0.3,
                    every arrow/.append style={
%                        -latex % Arrow tip, uncomment to get "arrows" instead of line segments
                },
            }\\
        }
    }
}
\begin{document}
\begin{questions}
\question[10]
Consider the differential equation
\begin{equation}
\dv{y}{x}(x,y)=\frac{x}{y^2},
\end{equation}
\begin{parts}
\part Sketch the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{center}
\def\length{sqrt(1+(x/y^2)^2)}
\begin{tikzpicture}[declare function={f(\x) = \x/\y^2;}]% Define which function we're using
\begin{axis}[ODE,
width=0.5\textwidth,
xmin=-5, xmax=5,
ymin=-5, ymax=5,
domain=-5:5, y domain=-5:5,
xtick={-5,-4,...,5}, ytick={-5,-4,...,5}
]
\addplot3 (x,y,0);
\addplot[red, samples=50, smooth, thick, no markers, domain=-5:5] {(3/2x^2-5)/abs(3/2x^2-5)abs(3/2x^2-5)^(1/3)}; %https://tex.stackexchange.com/questions/69411/pgfplots-cant-plot-some-usual-mathematical-functions
\end{axis}
\end{tikzpicture}
\end{center}
\vspace*{\stretch{1}}
\part Solve for the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{solution}
Separating variables,
\begin{align}
\dv{y}{x}&=\frac{x}{y^2} \
\int y^2 \dd{y} &= \int x\dd{x} \
\frac{1}{3}y^3 &= \frac{1}{2} x^2 + C
\end{align}
Substituting our initial condition,
\begin{align}
\frac{1}{3} &= 2 + C
\end{align}
so $C=-\frac{5}{3}$, which implies
\begin{align}
\frac{1}{3}y^3 &= \frac{1}{2} x^2 - \frac{5}{3} \
y^3 &= \frac{3}{2} x^2 -5 \
y &= \sqrt[3]{\frac{3}{2} x^2 -5}
\end{align}
\end{solution}
\vspace*{\stretch{1}}
\end{parts}
\end{questions}
\end{document}

I found a similar question experienced by another person - however the solution (restricting y domain) doesn't seem to help in this situation.

Thank you very much for your thoughts.

EDIT: One comment suggested removing the \\ from line 22 of the code. Doing so removes the error, but causes the slope field to display improperly:

Answer 1

There are many issues. One of them is that you divide by zero.

\documentclass{exam}
\printanswers
\usepackage{pgf,tikz}
\usepackage{pgfplots}
\usepackage{physics}
%Slope fields, from https://tex.stackexchange.com/questions/139064/how-to-draw-slope-fields-with-all-the-possible-solution-curves-in-latex
\pgfplotsset{ODE/.style={
    axis x line = middle,
    axis y line = middle,
    xlabel={$x$}, ylabel={$y$},
    axis equal image, % Unit vectors for both axes have the same length
    view={0}{90}, % We need to use "3D" plots, but we set the view so we look at them from straight up
    samples=21, % How many arrows?
    cycle list={    % Plot styles
            blue,
            quiver={
                u={1/\length}, v={f(x,y)/\length}, % End points of the arrows
                    scale arrows=0.3,
                    every arrow/.append style={
%                        -latex % Arrow tip, uncomment to get "arrows" instead of line segments
                },
            }\\
        }
    }
}
\begin{document}
\begin{questions}
\question[10]
Consider the differential equation
\begin{equation}
\dv{y}{x}(x,y)=\frac{x}{y^2},
\end{equation}
\begin{parts}
\part Sketch the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{center}
\def\length{sqrt(1+(x/y^2)^2)}
\begin{tikzpicture}[declare function={f(\x,\y) = \x/(\y+0.01)^2;}]% Define which function we're using
\begin{axis}[ODE,
width=0.5\textwidth,
xmin=-5, xmax=5,
ymin=-5, ymax=5,
domain=-5:5, y domain=-5:5,
xtick={-5,-4,...,5}, ytick={-5,-4,...,5}
]
\addplot3 (x,y,0);
\addplot[red, samples=50, smooth, thick, no markers, domain=-5:5] 
    {(3/2x^2-5)/abs(3/2x^2-5)abs(3/2x^2-5)^(1/3)}; %https://tex.stackexchange.com/questions/69411/pgfplots-cant-plot-some-usual-mathematical-functions
\end{axis}
\end{tikzpicture}
\end{center}
\vspace*{\stretch{1}}
\part Solve for the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{solution}
Separating variables,
\begin{align}
\dv{y}{x}&=\frac{x}{y^2} \
\int y^2 \dd{y} &= \int x\dd{x} \
\frac{1}{3}y^3 &= \frac{1}{2} x^2 + C
\end{align}
Substituting our initial condition,
\begin{align}
\frac{1}{3} &= 2 + C
\end{align}
so $C=-\frac{5}{3}$, which implies
\begin{align}
\frac{1}{3}y^3 &= \frac{1}{2} x^2 - \frac{5}{3} \
y^3 &= \frac{3}{2} x^2 -5 \
y &= \sqrt[3]{\frac{3}{2} x^2 -5}
\end{align}
\end{solution}
\vspace*{\stretch{1}}
\end{parts}
\end{questions}
\end{document}