Differentiate A = $\frac{\sqrt{(π^2r^6 + 9V^2)}}{r}$ with respect to r.
This function to the best of my knowledge is a multivariable function that can be differentiated partially, so I differentiated partially with respect to r and got $\frac{∂A}{∂r}$ = $\frac{6π^2r^5}{2r\sqrt{(π^2r^6 + 9V^2)}} - \frac{\sqrt{(π^2r^6 + 9V^2)}}{r^2}$
The original function can also be expressed as $(AR)^2 = π^2r^6 + 9V^2$ which is an implicit function and when differentiated with respect to r, I got $\frac{dA}{dr}$ = $\frac{(6π^2r^5) + (18V\frac{dv}{dr}) - (2A^2r)}{2r^2A}$
My question is, are these two derivatives the same? since the function was just expressed in another form.
Your partial differentiation of $$ A \ \ = \ \ \frac{\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }}{r} \ \ $$ is a little troubling, in that the problem statement doesn't indicate whether $ \ V \ $ is a function of $ \ r \ $ as well. Given that you want to compare your approaches, we should treat the derivative of $ \ V \ $ for both.
So if the area function is written as $ \ r^{-1}· ( \pi^2·r^6 \ + \ 9V^2 )^{1/2} \ \ , \ $ then the Product and Chain Rules give us
$$ \frac{dA}{dr} \ \ = \ \ \frac{6 \pi^2r^5 \ + \ \mathbf{18V·\frac{dV}{dr}} }{2·r·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} \ - \ \frac{\sqrt{\ \pi^2r^6 \ + \ 9V^2 }}{r^2} $$ $$ = \ \ \frac{3 \pi^2r^6 \ + \ 9·r·V\frac{dV}{dr} \ - \ ( \ \pi^2r^6 \ + \ 9V^2 \ ) }{r^2 ·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} \ \ = \ \ \frac{2 \pi^2r^6 \ + \ 9 r V\frac{dV}{dr} \ - \ 9V^2 }{r^2 ·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} \ \ , $$ upon placing the entire expression over a common denominator.
Differentiating implicitly with respect to $ \ r \ \ , \ $ the relation $ \ A^2r^2 \ = \ \pi^2r^6 \ + \ 9V^2 \ $ produces
$$ 2Ar^2 · \frac{dA}{dr} \ + \ 2A^2r \ \ = \ \ 6 \pi^2r^5 \ + \ 18V\frac{dV}{dr} \ \ \Rightarrow \ \ \frac{dA}{dr} \ \ = \ \ \frac{ 3 \pi^2r^5 \ + \ 9V\frac{dV}{dr} \ - \ A^2r}{A·r^2} \ \ . $$
Substituting the original expression for $ \ A \ $ leads to
$$ \frac{dA}{dr} \ \ = \ \ \frac{ ( \ 3 \pi^2r^5 \ + \ 9V\frac{dV}{dr} \ ) \ - \ \frac{ \pi^2r^6 \ + \ 9V^2 }{r^2} · r }{r·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} $$ $$ = \ \ \frac{ ( \ 3 \pi^2r^5 \ + \ 9V\frac{dV}{dr} \ )·r \ - \ \frac{ \pi^2r^6 \ + \ 9V^2 }{r^2} · r^2 }{r^2·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} $$ $$ = \ \ \frac{ 3 \pi^2r^6 \ + \ 9rV\frac{dV}{dr} \ - \ ( \ \pi^2r^6 \ + \ 9V^2 \ ) }{r^2·\sqrt{ \ \pi^2r^6 \ + \ 9V^2 }} \ \ , $$ which gives the same result as found above.
So the total derivative (as remarked by Guangliang) found from either method is the same. If $ \ V \ $ is independent of $ \ r \ \ , \ $ the term $ \ 9rV\frac{dV}{dr} \ $ will be absent in both methods.