Answer
Local minimum and maximum at $(-1,\pm 1)$; Saddle point at $(-1,0),(1,\pm 1)$
Work Step by Step
Given: $f(x,y)=3x-x^3-2y^2+y^4$
This yields $f_x(x,y)=3-3x^2$ and $f_y(x,y)=4y^3-4y$ and $(x,y)=(-1,\pm 1)$ or $(1,0)$.
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.
Hence,
Local minimum and maximum at $(-1,\pm 1)$; saddle point at $(-1,0),(1,\pm 1)$.
