What are the critical point(s) of h(x)=(9-4x^2)^(1/2) ?
Well if you were going to do it by hand, do it this way...
$\displaystyle \displaystyle \begin{align*} h &= \left( 9 - 4x^2 \right)^{\frac{1}{2}} \end{align*}$
Let $\displaystyle \displaystyle \begin{align*} u = 9 - 4x^2 \end{align*}$ so that $\displaystyle \displaystyle \begin{align*} h = u^{\frac{1}{2}} \end{align*}$.
Then $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = -8x \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \frac{dh}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\left( 9 - 4x^2 \right) ^{\frac{1}{2}}} \end{align*}$
So $\displaystyle \displaystyle \begin{align*} \frac{dh}{dx} = \frac{-8x}{2 \left( 9 - 4x^2 \right)^{\frac{1}{2}} } = \frac{-4x}{\left( 9 - 4x^2 \right)^{\frac{1}{2}}} \end{align*}$
Yes, you are correct that when $\displaystyle \displaystyle \begin{align*} \frac{dh}{dx} = 0 \end{align*}$ then $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$. What is the value of h when x = 0?