As an alternative to @Riccardo.Alestra's fine answer,
and with a discussion of proper treatment of critical points
(to justify that the solution is indeed a global minimum)...
We wish to minimize $\pi(r^2+2rh)$ subject to $\pi r^2h=V$. Without preference for either $r$ or $h$, we could proceed using differentials. The constraint becomes a relation between $dr,dh$:
$$
0 = dV = \pi \cdot d\left(r^2h\right)
$$
or, dispensing with the multiples of $\pi$ and then $r$,
$$
0 = r^2 dh + 2rh \, dr \implies
$$
$$
0 = r \, dh + 2h \, dr \implies
$$
$$
\frac{dr}{dh}=-\frac12\frac{r}{h}
\qquad
\text{or}
\qquad
\frac{dh}{dr}=-2\frac{h}{r}
\,.
$$
Now we turn to the objective function, also
dispensing with the constant multiples of $\pi$ and then $2$:
$$
\frac{A}{\pi} = r^2+2rh
$$
$$
\eqalign{
0
&= d\left( r^2+2rh \right) \\
&= 2r\,dr + 2\left( r\,dh + h \, dr \right)
\implies\\
0
&= r\,dr + r\,dh + h \, dr \\
&= \left(r+h\right)\,dr + r\,dh
}
$$
At this point, we use one of the two equivalent differential ratios above:
$$
\eqalign{
0
&= r+h + r\,\frac{dh}{dr} \\
&= r+h + r\,\left(-2\frac{h}{r}\right) \\
&= r+h -2 r\,\frac{h}{r} \\
&= r+h -2 h \\
&= r-h \\\\
&\iff\qquad r=h
}
$$
Putting this back into the constraint
(and being forced to prefer one variable, say $r$),
we obtain
$$
\eqalign{
V &= \pi r^3 = \pi h^3 \\
r &= h = \left(\frac{V}{\pi}\right)^{1/3} \\
}
$$
Lastly, we need to ensure that this is a global minimum
and not a local minimum or global or local maximum.
To see this, we either need the second derivative of our objective function
$f$ or else a numberline sketch of the sign of $f\,'$ for $r,h>0$ (satisfying the constraint, which should also be graphed to see the inverse relationship).
Recall that our objective function
$$
f(r)=\pi\left(r^2+2rh\right)
$$
has derivative
$$
f\,'(r)=\pi r\left(r-h\right)
$$
which is negative for $r\in(0,h)$ and positive for $r > h$,
so that $r=h$ is indeed the global minimum.
One can also, of course, compute
$$
\eqalign{
f\,''(r)
&= \pi \, \frac{d}{dr} \left( r^2 - rh \right) \\
&= \pi \, \left( 2r - h - r \, \frac{dh}{dr} \right) \\
&= \pi \, \left( 2r + h \right) \implies \\
\Bigl. f\,''(r) \Bigr|_{r=h} &= 3\pi r > 0
}
$$
which shows that $r=h$ is at least a local minimum,
but we must observe that $f\,''>0$ for all $r,h>0$
(i.e. that $f$ is strictly concave)
to conclude that it is in fact a global minimum.