# How do you implicitly differentiate 2=x^2lny-y?

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y \ln \left(y\right)}{y - {x}^{2}}$
Here you need to consider $y$ as a function of $x$ and so differentiate it as well.
$0 = 2 x \ln \left(y\right) + {x}^{2} \cdot \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} - 1 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x \ln \left(y\right)}{{x}^{2} / y - 1} = \frac{2 x y \ln \left(y\right)}{y - {x}^{2}}$