A doubt over a differentiation problem. The question is:

Find the equation of the tangent to the curve $y^2=8x$ at a point $(x_0,y_0)$.

My teacher's approach :

Differentiate the equation. we get $2y\cdot \dfrac {dy}{dx}=8 $
which gives us : $\dfrac {dy}{dx}=\dfrac4y$.
At a point $(x_0,y_0)$, slope of the tangent =$\dfrac4{y_0}$
So the equation of the tangent is $\dfrac{y-y_0}{x-x_0}=\dfrac4{y_0}$

What bothers me here is the differentiation. According to wikipedia :

the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point.

But, $y^2=8x$ is NOT a function. Then, how can one find the derivative?
This bothers me a lot.. Can someone help me? Thank you.

Forget about $y$ and start writing $f(x)^2=8x$. Now you do have a function ( it's just a matter of which symbols you are using ).
Derive this function in respect to $x$, in the left side you have $2f '(x)f(x)$, using the chain rule, and in the right side you have $8$. Both sides are equal, so you get $2f '(x)f(x)=8$, therefore $f '(x)=4/f(x)$.
