Answer
Local maximum at $f(0,0)=1$
Work Step by Step
Given: $f_x(x,y)=\dfrac{-2x}{3(x^2+y^2)^{2/3}}=0, f_y(x,y)=\dfrac{-2y}{3(x^2+y^2)^{2/3}}=0$
Simplify the given two equations.
This implies that $x=0,y=0$
Critical point: $(0,0)$
In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum.
2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum.
3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point.
$D=f_{xx}f_{yy}-f^2_{xy}=1-\sqrt [3] {x^2+y^2}\implies D=1-\sqrt [3] {x^2+y^2}\gt 0$ and $f_{xx}=-\dfrac{-8}{15} \lt 0$
Hence, Local maximum at $f(0,0)=1$
