How to plot a figure according to two equations?

Question

There are two equations like this: $a=f(x,y)$ and $z=g(x,y)$, where $a$ is a constant. In principle, one can solve $a=f(x,y)$ for $y$ as $y=k(x)$ and substitute it into the second equation to obtain $z=g(x,k(x))=h(x)$. I want to plot the curves of $z=h(x)$. However, the first equation is too complicated to solve $y$ as function of $(x,a)$ and the second equation is also very complicated. In this case, how can I get the curves?

e.g. the first equation as

1=(-1 + 2 x^2 y)/((x + x^3 y) Sqrt[ 1 + y - Sqrt[1 + 1/(x y^2) + (2 x)/y] y]) 

and the second equation as

z=(-4 + x^2 (2 - Sqrt[1 + 1/(x y^2) + (2 x)/y]) y^2 + (x - x Sqrt[1 + 1/(x y^2) + (2 x)/y]) y^3 + y (-3 + 2 Sqrt[1 + 1/(x y^2) + (2 x)/y] + 2 x Sqrt[1 + 1/(x y^2) + (2 x)/ y] (1 - 2 Sqrt[1 + y - Sqrt[1 + 1/(x y^2) + (2 x)/y] y])))/(6 x Sqrt[1 + 1/(x y^2) + (2 x)/y] y Sqrt[1 + y - Sqrt[1 + 1/(x y^2) + (2 x)/y] y])`

Answer 1

One possible approach. As murray points out in the comment, actually $y$ is not determined uniquely from $x$, so not to lose information it seems better to look at the whole thing in 3D. But first I believe it is necessary to get rid of radicals. Denoting Sqrt[1 + 1/(x y^2) + (2 x)/y] by $u$ and Sqrt[ 1 + y - Sqrt[1 + 1/(x y^2) + (2 x)/y] y] by $v$ one obtains algebraic equations $u^2 x y^2-2 x^2 y-x y^2=1$, $-u y+y+1=v^2$ for them, while the original ones become $(x+x^3y)v=2x^2y-1$ and $z=\frac{-4 u v x y-u x^2 y^2-u x y^3+2 u x y+2 u y+2 x^2 y^2+x y^3-3 y-4}{6 u v x y}$.

It is straightforward to then eliminate $u$ and $v$ obtaining $f(x,y)=1$ and $z=g(x,y)$ where $$ f(x,y)=2 x^{13} y^5-2 x^{12} y^5-x^{12} y^4+9 x^{11} y^4+8 x^{10} y^5-4 x^{10} y^3+16 x^9 y^3-8 x^8 y^4-4 x^8 y^3-6 x^8 y^2+14 x^7 y^2+26 x^6 y^3-14 x^6 y^2-4 x^6 y+6 x^5 y-28 x^4 y^2-6 x^4 y-x^4+x^3+10 x^2 y+2 x^2 $$ and \begin{align*} g&(x,y)=\\ &\left(x^{10} y^5-x^{10} y^4-x^9 y^5+2 x^9 y^4+2 x^9 y^3-2 x^8 y^4-13 x^8 y^3+4 x^7 y^5-3 x^7 y^4-2 x^7 y^3+6 x^7 y^2+12 x^6 y^3-21 x^6 y^2-3 x^5 y^3+6 x^5 y^2+6 x^5 y-53 x^4 y^2-7 x^4 y-3 x^3 y^3-x^3 y^2+8 x^3 y+2 x^3+34 x^2 y+2 x^2+x y^2-2 x-6\right)/\left(6 \left(2 x^2 y-1\right) \left(x^6 y^3+x^6 y^2-2 x^4 y^2+2 x^4 y+5 x^2 y+x^2-1\right)\right) \end{align*} Then

ContourPlot3D[{-6 - 2 x + 2 x^2 + 2 x^3 + 34 x^2 y + 8 x^3 y - 
7 x^4 y + 6 x^5 y + x y^2 - x^3 y^2 - 53 x^4 y^2 + 6 x^5 y^2 - 
21 x^6 y^2 + 6 x^7 y^2 - 3 x^3 y^3 - 3 x^5 y^3 + 12 x^6 y^3 - 
2 x^7 y^3 - 13 x^8 y^3 + 2 x^9 y^3 - 3 x^7 y^4 - 2 x^8 y^4 + 
2 x^9 y^4 - x^10 y^4 + 4 x^7 y^5 - x^9 y^5 + x^10 y^5 - 
6 (-1 + 2 x^2 y) (-1 + x^2 + 5 x^2 y + 2 x^4 y - 2 x^4 y^2 + 
   x^6 y^2 + x^6 y^3) z == 0,
2 x^2 + x^3 - x^4 + 10 x^2 y - 6 x^4 y + 6 x^5 y - 
4 x^6 y - 28 x^4 y^2 - 14 x^6 y^2 + 14 x^7 y^2 - 6 x^8 y^2 + 
26 x^6 y^3 - 4 x^8 y^3 + 16 x^9 y^3 - 4 x^10 y^3 - 8 x^8 y^4 + 
9 x^11 y^4 - x^12 y^4 + 8 x^10 y^5 - 2 x^12 y^5 + 2 x^13 y^5 == 1},
{x, -7, 7}, {y, -7, 7}, {z, -7, 7}, PlotPoints -> 70, Mesh -> None]

shows that indeed there are several branches of $y$ above the $(x,z)$ plane to sort out

(The $f(x,y)=1$ surface is the blue one)

PS Just noticed the comment by Rahul with a link to a similar but much better solution. Will still leave mine as it is since it has been upvoted :P