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.