Consolidating my comments (so that they can be cleaned up): This is a misunderstanding.
A twice (continuously!) differentiable function $f:\mathbb{R}^n\to \mathbb{R}$ is convex if and only if the Hessian $\nabla^2 f(x)\in\mathbb{R}^{n\times n}$ is positive semi-definite at every $x\in \mathbb{R}^n$. (This definition makes sense since the Hessian is symmetric by Schwarz' theorem if the second derivatives are continuous.) This is sometimes written as
$$\nabla^2 f(x) \succeq 0 \qquad\text{for all } x\in\mathbb{R}^n$$
(and more rarely -- since it can lead to misunderstandings -- as $\nabla^2 f(x)\geq 0$).
As @nicoguaro points out in his answer, this is equivalent to the condition that all eigenvalues of $\nabla^2 f(x)$ -- as a function of $x$ -- are nonnegative for every $x\in \mathbb{R}^n$. An equivalent (and often easier to verify, especially for large $n$) condition is that
$$d^T\nabla^2 f(x)d \geq 0 \qquad\text{for all } d\in\mathbb{R}^n \text{ and }x\in\mathbb{R}^n.$$
(This condition is also easier to work with if you want to rule out convexity: It's sufficient to find a single $d$ such that $d^T \nabla^2 f(x) d<0$.)
In your example (with $x_1 = x$ and $x_2 = y$), this would yield
$$
\begin{aligned}
\begin{pmatrix} d_1 & d_2 \end{pmatrix}
\begin{pmatrix}
\frac{x_2^2}{(x_1^2 + x_2^2)^\frac{3}{2}} &
\frac{-x_1\,x_2}{(x_1^2 + x_2^2)^\frac{3}{2}} \\
\frac{-x_1\,x_2}{(x_1^2 + x_2^2)^\frac{3}{2}} &
\frac{x_1^2}{(x_1^2 + x_2^2)^\frac{3}{2}}
\end{pmatrix}
\begin{pmatrix} d_1 \\ d_2 \end{pmatrix}
&=
\frac{1}{(x_1^2 + x_2^2)^\frac{3}{2}}\left(d_1^2x_2^2 - 2 d_1 x_1x_2d_2 + d_2^2x_1^2\right)\\
&= \frac{1}{(x_1^2 + x_2^2)^\frac{3}{2}}\left(d_1x_2-d_2x_1\right)^2\\
&\geq 0
\end{aligned}
$$
for all $x,d\in\mathbb{R}^n$. Hence, $f$ is convex.
