Suppose we have the equation $V = \frac{1}{3}\pi r^2h$. Find $\frac{dr}{dh}$.
[Chain Rule]
We have $\frac{dV}{dh}=\frac{dV}{dr}\cdot\frac{dr}{dh}$. $$ \begin{cases} \frac{dV}{dh}=\frac{1}{3}\pi r^2\\ \frac{dV}{dr}=\frac{2}{3}\pi rh \end{cases} \implies \frac{1}{3}\pi r^2 = \frac{2}{3}\pi rh\cdot\frac{dr}{dh} \implies r=2h\cdot\frac{dr}{dh} \implies \frac{dr}{dh}=\frac{r}{2h} $$
[Implicit Differentiation]
$$ \begin{align*} \frac{d}{dh}V&=\frac{\pi}{3}\frac{d}{dh}\big(r^2 h\big)\\ 0&=\frac{\pi}{3}\big(r^2\cdot\frac{d}{dh}h + h\frac{d}{dh}r^2\big)\\ 0&=r^2+2\cdot h\cdot r\frac{dr}{dh}\\ \frac{dr}{dh}&=-\frac{r^2}{2\cdot h\cdot r}=-\frac{r}{2h} \end{align*} $$
Why does the solution using the chain rule method have a different sign compared to the one using implicit differentiation? Did I make any mistake?
There's a few things going on here. It's important to distinguish between $V(r,h)$ the function and $V$ the constant.
First, your chain rule is wrong (for what you are asking). It should be: $$\dfrac{d V}{d h} = \dfrac{\partial V(r,h)}{\partial r} \dfrac{d r}{d h} + \dfrac{\partial V(r,h)}{\partial h} \dfrac{d h}{d h}$$ Notice that the left hand side is $V$ the constant, which when you did implicit differentiation you set equal to zero. Then we have \begin{align} \dfrac{d V}{d h} &= \dfrac{\partial V(r,h)}{\partial r} \dfrac{d r}{d h} + \dfrac{\partial V(r,h)}{\partial h} \dfrac{d h}{d h}\\ 0 &= (\dfrac{2}{3} \pi r h) \dfrac{d r}{d h} + \dfrac{1}{3} \pi r^2 (1)\\ \implies \dfrac{d r}{d h} &= -\dfrac{r}{2h} \end{align}
Now chain rule gives the same answer as implicit differentiation.
To give a bit more intuiton.