How to derive the equation of tangent to an arbitrarily point on a ellipse? 
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$

I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:)
Thanks in advance!
Edit:
Solving for $\frac{dy}{dx}$ and putting it into point slope gives me:
$\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$ 
Then we get:
$y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,*y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$
Which looks close, but not exactly the expression i wanted. Where is the error?

To show that the given line is tangent to the ellipse, one can show that the system of equations
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \tag{I} \\
\frac{xx_0}{a^2} + \frac{yy_0}{b^2} &= \tag{II} 1
\end{align}
with $x$, $y$ variables has exactly one solution—namely $(x_0, y_0)$. We further assume that $(x_0, y_0)$ lies on the ellipse, i.e.,
$$ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1. \tag{III} $$
Taking $(\mathrm I) - 2(\mathrm{II}) + (\mathrm{III})$ produces
$$ \frac{x^2 - 2xx_0 + x_0^2}{a^2} + \frac{y^2 - 2yy_0 + y_0^2}{b^2} = 0 $$
or
$$ \left(\frac{x-x_0}{a}\right)^2 + \left(\frac{y-y_0}{b}\right)^2 = 0. $$
The only way for the sum of two squares to be zero is that $x=x_0$ and $y=y_0$, hence this is indeed the only solution as desired.
