Answer
Minimum values: $( \dfrac{1}{\sqrt2}, -\dfrac{1}{\sqrt2}),( - \dfrac{1}{\sqrt2}, \dfrac{1}{\sqrt2})=-\dfrac{1}{4}$
Saddle point:$(0,0)$
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 a not a local minimum and local maximum or, a saddle point.
As we are given that $f(x,y)=x^2+y^4+2xy$
Here, $f_x=2x+2y, f_y=4y^3+2x$
After simplifications, we get $x=\dfrac{1}{2}, \pm \dfrac{1}{\sqrt2}$ and $y= \mp \dfrac{1}{\sqrt2}$
For $(x,y)=( \pm \dfrac{1}{\sqrt2}, \mp \dfrac{1}{\sqrt2})$
$D(0,0)=8 \gt 0$ ; and $f_{xx}( \pm \dfrac{1}{\sqrt2}, \mp \dfrac{1}{\sqrt2})=2 \gt 0$
$D(0,0)=-4 \lt 0$; saddle points
Also, Plug $f(x,y)=f( \dfrac{1}{\sqrt2}, -\dfrac{1}{\sqrt2})$ in the given function , we get $-\dfrac{1}{4}$
Hence,
Minimum values: $( \dfrac{1}{\sqrt2}, -\dfrac{1}{\sqrt2}),( - \dfrac{1}{\sqrt2}, \dfrac{1}{\sqrt2})=-\dfrac{1}{4}$
Saddle point:$(0,0)$
