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?

Trying plugging in those values of $x$, you'll notice that the Hessian of $f$ doesn't change for any value of $k$.
You could consider studying the critical points of $f$ only in a $\{(x,y)\mid -\frac{\pi}2<x<\frac{\pi}2\}$ strip first, and deduce later that all the other critical points are just translated copies of these one.
