Answer
$f(\dfrac{1}{3},\dfrac{-2}{3})$ has local minimum at $-\dfrac{1}{3}$
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.
For the given question, we have $f_x(x,y)=2x+y$ and $f_y(x,y)=x+2y+1$ gives $(x,y)=(\dfrac{1}{3},\dfrac{-2}{3})$.
For $(x,y)=(\dfrac{1}{3},\dfrac{-2}{3})$
$D(\dfrac{1}{3},\dfrac{-2}{3})=3 \gt 0$ and $f_{xx}(\dfrac{1}{3},\dfrac{-2}{3})=2 \gt 0$
This implies that $f(\dfrac{1}{3},\dfrac{-2}{3})$ has local minimum at $-\dfrac{1}{3}$
