Answer
$(a)$
Local maximum: $(1,\displaystyle \frac{1}{3})$
local minimum: $(0,0).$
$(b)$
Absolute maximum: none.
Absolute minimum: $(0,0)$
$(c)$
see graph
Work Step by Step
$g(x)=\displaystyle \frac{x^{2}}{4-x^{2}},\quad x\in(-2,1]$
$g'(x)=\displaystyle \frac{2x(4-x^{2})-x^{2}(-2x)}{(4-x^{2})}=\frac{8x}{(2-x)^{2}(2+x)^{2}}$
$g'(x)$ is undefined for $ x=\pm 2\qquad$ ... not critical points (not in the domain).
$g'(x)=0$ for $ x=0\qquad$ ... critical point.
$\left[\begin{array}{ccccccc}
interval & ( & (-2,0) & & (0,1) & ]\\
t & -2 & -1 & 0 & 0.5 & 1\\
f'(t) & & -0.89 & & 0.284 & \\
f(t) & & \searrow & 0 & \nearrow & 1/3\\
& & & & &
\end{array}\right]$
