<s>[INST] 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?
 [/INST] We first differentiate the ellipse equation w.r.t. $x$  to find the slope of the tangent $m$,
$$\dfrac{2x_0}{a^2}+\dfrac{2y_0m}{b^2}=0$$
$$m=-\dfrac{x_0}{y_0}\cdot\dfrac{b^2}{a^2}$$
Thus, the tangent is given by
$$\color{darkgreen}{y-y_0}=\color{blue}{-}\dfrac{\color{blue}{x_0}}{\color{darkgreen}{y_0}}\cdot\dfrac{\color{darkgreen}{b^2}}{\color{blue}{a^2}}\color{blue}{(x-x_0)}$$
$$\color{darkgreen}{\dfrac{yy_0}{b^2}-\dfrac{y_0^2}{b^2}}=\color{blue}{\dfrac{x_0^2}{a^2}-\dfrac{xx_0}{a^2}}$$
$$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}=\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=1$$
as desired.
Hope this helps. :)
</s>