Answer
Saddle points at $(\dfrac{\pi}{2}+p \pi,0)$, where $p$ is any positive integer.
Work Step by Step
Given: $f(x,y)=y \cos x$
This yields $f_x(x,y)=-y\sin x$ and $f_y(x,y)=\cos x$ gives no values of $x=\dfrac{\pi}{2}+n \pi, y=0$.
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
ii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
iii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
For $(x,y)=(\dfrac{\pi}{2}+n \pi,0)$, we have $D(\dfrac{\pi}{2}+n \pi,0)=- \sin^2 x \lt 0$
This shows that Saddle points at $(\dfrac{\pi}{2}+p \pi,0)$, where $p$ is any positive integer.
