Answer
There is a saddle point at $(n \pi, 0)$ where, $n=0,1,2 \dots$
Work Step by Step
$y \cdot \cos x=f_{x}(x, y)$
$\sin x=f_{v}(x, y)$
At critical points,
$y \cdot \cos x=0$
$\sin x=0$
$n=x \pi$ where, $n=0,1,2 \dots$
and $y=0$
$-y . \sin x=f_{x x}(x, y)$
$f_{y y}(x, y)=0$
$f_{x y}(x, y)=\cos x$
at $(n \pi, 0)$
$f_{x x}(n \pi, 0)=0$
$f_{y y}(n \pi, 0)=0$
$f_{x}(n \pi, 0)=\left\{\begin{array}{ll}1 & \text { if } n \text { is } 0 \text { or even } \\ -1 & \text { if } n \text { is odd }\end{array}\right.$
$D(n \pi, 0)=-1<0$ for all value of $n$
So, there is a saddle point at $(n \pi, 0)$ where $n=0,1,2 \dots$.
