Answer
Absolute maximum: $f(1,\sqrt 2)=2$
Absolute minimum: $f(0,0)=0$
(and also along $(0,y)$ and $(x,0)$)
Work Step by Step
To calculate the critical points, we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$.
Thus, $f_x=y^2,f_y=2xy$
This yields, $x=0,y=0$ and $f(0,0)=0$
From the the given boundary $x^2+y^2=3$, we have $y=\sqrt {3-x^2}$
$f(x)=x(3-x^2)$ and $f'(x)=3-3x^2$
so, $x=1$ and $y=\sqrt {3-x^2} =\sqrt 2$
Thus, Critical point: $(1,\sqrt 2)$
At $(x,y)=(1,\sqrt 2)$
$f(1,\sqrt 2)=2$
Therefore, it has minimum and maximum values at $0,2$
Hence,
Absolute maximum: $f(1,\sqrt 2)=2$
Absolute minimum: $f(0,0)=0$
(and also along $(0,y)$ and $(x,0)$)
