Study the local minimum, maximum points or saddle points for the function $f(x,y)$ where $f(x,y)=xy^2+\sin(x)$ in the region $A = \{(x,y)\mid x\ge 0\}$
my attempt:
$\left(\frac{\partial f}{\partial x}\right)(x,y)= y^2+\cos(x) $
$\left(\frac{\partial^2 f}{\partial x^2}\right)(x,y)= -\sin(x) $
$\left(\frac{\partial f}{\partial y}\right)(x,y)= 2 y x $
$\left(\frac{\partial^2 f}{\partial y^2}\right)(x,y)= 2 x $
$\left(\frac{\partial f}{\partial x \partial y}\right)(x,y)= 2 y $
to find the critical points :
at $\left(\frac{\partial f}{\partial x}\right)(x,y)= 0 $
$y^2+\cos(x) =0 $ then $ x=\left(\frac{\pi}{2}\right)+2\pi n$ , $ x=\left(\frac{3\pi}{2}\right)+2\pi n$ and $y=0$
at $\left(\frac{\partial f}{\partial y}\right)(x,y)= 0 $
$2yx=0 $ then $ x=0 , y=0$
if $\left(\frac{\partial^2 f}{\partial x^2}\right)(x,y).\left(\frac{\partial^2 f}{\partial y^2}\right)(x,y)-\left(\frac{\partial f}{\partial x \partial y}\right)^2(x,y) \gt 0 $ it's a maximum or minimum
if $\left(\frac{\partial^2 f}{\partial x^2}\right)(x,y) \gt 0 $ then it's a minimum point
if $\left(\frac{\partial^2 f}{\partial x^2}\right)(x,y) \lt 0 $ then it's a maximum point
if $\left(\frac{\partial^2 f}{\partial x^2}\right)(x,y).\left(\frac{\partial^2 f}{\partial y^2}\right)(x,y)-\left(\frac{\partial f}{\partial x \partial y}\right)^2(x,y) \lt 0 $ it's a saddle point
My problem is in the critical points. I'm confused.