Finding Equation of tangent to the curve $xy^2=25$ at point $(5,5)$ 
Find equation of the tangent to the curve $xy^2=25$ at point $(5,5)$

My Working
I tried this sum using two different approaches. But I'm getting two different answers. I wonder if any is correct. Can you please guide me on this
First approach;
use product rule to differentiate expression as it is to find the $\frac{dy}{dx}$ gives me,
$x.2y\frac{dy}{dx}+y^2.1=0$
$\frac{dy}{dx}=-\frac{y}{2x}$
Finally Im getting the equation as, $y-5=-\frac{1}{2}(x-5)$
Second approach;
differentiate function explicitly after making $y^2$ the subject,
$y^2=\frac{25}{x}$
$2y\frac{dy}{dx}=-\frac{25}{x^2}$
$\frac{dy}{dx}=-\frac{25}{2yx^2}$
$y-5=-\frac{1}{10}(x-5)$
Thank you in advance!

You did the differentiation right.
$$\frac{dy}{dx} = - \frac{y}{2x} = - \frac{25}{x^2y}$$
Note that if $xy^2 = 25$, then $- \frac{25}{2x^2y} = - \frac{25y}{2x^2y^2} = - \frac{25y}{2x(xy^2)} = - \frac{25y}{2x(25)} = - \frac{y}{2x}$, so the two expressions are equivalent.
The problem is that the question is ill-formed: If $x = y = 5$, then $xy^2 = 125 \ne 25$.  So with two contradictory values of $xy^2$, of course the math won't be consistent.
