Given the function: $f(x,y) = \cos y\cdot e^x$ Question is to determine the nature of its critical points. I've calculated the partial derivatives as following :
$\frac{\partial f}{\partial x} (x,y)=\cos y\cdot e^x$
$\frac{\partial f}{\partial y} (x,y)=-\sin y\cdot e^x$
None of these can equal to zero does that mean that the function doesn't have a critical point or is there another method?
Function $f$ is differentiable in $\mathbb{R}^2$. By definition, its critical points satisfy the stationarity conditions you mentioned. Since those conditions are not met at any point $(x,y)$, we conclude that $f$ has no critical points.