critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$ so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2}\\ -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2} & \frac{8\,y^2\,\sin\left(x\right)}{{\left(y^2+1\right)}^3}-\frac{2\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
My first problem: What $x$ and $y$ do I plug in the hessian-matrix? If I solve $f'(x,y)$ for $(0,0)$ I get:
$x = \pi \cdot (k+1/2)$ or $x = \pi \cdot k$
Now, do I have to substitute these values for $x$ and solve for $y$? Seems like a huge case-differentiation to me.
...There must be an easier way, right?

Note that for:
$$f'(x,y) = (0,0) \iff \frac{cos(x)}{y^2+1} = 0 \quad \wedge \quad \frac{y\operatorname{sin}(x)}{(y^2+1)^2} = 0 $$
The first condition can only be achieved if $x\in \frac{\pi}{2}+\pi\mathbb{Z}$, and since cos and sin are linearly independent, then the only way they can both be 0 at the same time is for y to be equal to 0. Thus, the critical points are of the form $\lbrace \pi/2 + k\pi: k\in\mathbb{N}\rbrace\times\lbrace 0\rbrace$. Now you can just plug them in the Hessian
