Answer
Saddle points are at $(\dfrac{\pi}{2}+n \pi,0)$, where $n$ is any positive integer.
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. 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.
2.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.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum, local maximum, or a saddle point.
From the given question, we have $f_x(x,y)=-y\sin x$ and $f_y(x,y)=\cos x$ gives no values of $x=\dfrac{\pi}{2}+n \pi, y=0$.
For $(x,y)=(\dfrac{\pi}{2}+n \pi,0)$
$D(\dfrac{\pi}{2}+n \pi,0)=- \sin^2 x \lt 0$
Hence, saddle points are at $(\dfrac{\pi}{2}+n \pi,0)$, where $n$ is any positive integer.
