Define when $y$ is a function of $x$ Hello guys I want to make  sure myself  in determming when is  $y$ as function of   $x$,so for this,  let us consider following question. If the equation of circle is given by  this
$$x^2+y^2=25$$
and question is  find the equation of tangent  of circle at point $(3,4)$,then it is clear that we need  calculate the derivative  and also we can  express $ y$ as
$$y=\sqrt{25-x^2}$$
but, let us consider following situation 

$$x^3+y^3=6\cdot x \cdot y$$

it's name is The Folium of Descartes  and my question is  the same : calculate  the equation of the tangent of this  folium at point $(3,4)$.
I know that  somehow expressing  $y$ as a function of $x$ is difficult  but could't we do it by some even long mathematical manipulation? So what  is the strict explanation when is $y$ a  function of $x$ and when is it not? I need to be    educated in this  topic and please explain me ways  of this,thanks a lot.
 A: The equation $$
\tag{1}x^3+y^3=6xy
$$does define $y$ as a function of $x$ locally (or, rather, it defines $y$ as a function of $x$ implicitly).  Here, it is difficult to write the defining equation  as $y$ in terms of $x$.  But, you don't have to do that to evaluate the value of the derivative of $y$.
[edit] The point $(3,4)$ does not satisfy   equation (1); so there is no tangent line at this point. 
Let's, instead,  consider the point $(3,3)$, which does satisfy equation (1):
To find the slope of the tangent line at $(3,3)$, you need to find $y'(3)$. To find this, first
  implicitly differentiate both sides of the defining equation for $y$ 
 (equation (1)). This gives 
$$
{d\over dx} (x^3+y^3)={d\over dx} 6xy
$$
So,  using the chain and product rules:
$$
3x^2+3y^2 y' =6y+6x y'.
$$ 
When $x=3$ and $y=3$, you have
$$
3\cdot 3^2+3\cdot 3^2 y'(3)=6\cdot 3+6\cdot 3\cdot y'(3).
$$
Solve this for $y'(3)={3\cdot 3^2-6\cdot 3\over6\cdot3 -3\cdot3^2}=-1.$
Now you can find the equation of the tangent line since you know the slope and that the point  $(3,3)$ is on the line.

Generally any "nice" equation in the variables $x$ and $y$ will define $y$ as a function of $x$ in some neighborhood of a given point. Given the $x$ value, the corresponding $y$ value is the solution to the equation ("the" solution in a, perhaps small, neighborhood of the point). 

Of course, sometimes it is extremely difficult (if not impossible) to find an explicit form of the function; that is, of the form $y=\Phi(x)$ for some expression $\Phi(x)$.  In these cases, to find the derivative of $y$, you have to use the approach above.
