Answer
Local maximum at $f(0,0)=1$
Work Step by Step
Since, we have $f_x(x,y)=\dfrac{-2x}{3(x^2+y^2)^{2/3}}=0, f_y(x,y)=\dfrac{-2y}{3(x^2+y^2)^{2/3}}=0$
There are no solutions to the above equations, but the partial derivatives are undefined at:
$x=0,y=0$
Thus the critical point is: $(0,0)$
We see that $f(x,y)\le 1$ for all points except $(0,0)$.
Thus, local maximum at $f(0,0)=1$
