Find the derivative of the following equation.. I have a question in my manual and I am not able to answer it, I'd appreciate some help please.
Find $ \dfrac{dy}{dx} $  if  $ 2x^2y + 3xy^2 = 6 $
I'm confused with = 6.. Thanks !
 A: Hint: This is almost solution to your question:
If equation is $F(x,y)= c$ for some constant $c$, then 
$\frac{d}{dx}(F(x,y))= 0$.  Use derivative function is linear and 
If somewhere you come across the term $x^ny^m$ we have $$\frac{d}{dx}x^ny^m= x^n.\frac{d}{dx}y^m+ y^m\frac{d}{dx}x^n$$
and $\frac{d}{dx}y^m= my^{m-1}\frac{dy}{dx}$
A: Consider $y$ as a function of $x$ defined implicitly by 
$$\begin{equation*}
2x^{2}y+3xy^{2}=6.
\end{equation*}$$
The derivatives of both sides should be equal. The derivative of the RHS is $0$, because the derivative of a constant is $0$. As for the derivative of the LHS, by the sum and product rules, is given by
$$\begin{eqnarray*}
\frac{d}{dx}\left( 2x^{2}y+3xy^{2}\right)  &=&\frac{d}{dx}\left(
2x^{2}y\right) +\frac{d}{dx}\left( 3xy^{2}\right)  \\
&=&2\frac{d}{dx}\left( x^{2}y\right) +3\frac{d}{dx}\left( xy^{2}\right). 
\end{eqnarray*}$$
Therefore
$$
\begin{equation*}
2\frac{du}{dx}+3\frac{dv}{dx}=0,
\end{equation*}$$
where $u=x^{2}y$ and $v=xy^{2}$. Compute the total derivatives to evaluate $du/dx$ and $dv/dx$. They can be expressed in terms of the partial derivatives and the derivative $dy/dx$ you want to find as follows. 
$$
\begin{eqnarray*}
\frac{du}{dx} &=&\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}
\frac{dy}{dx}=2xy+x^{2}\frac{dy}{dx} \\
\frac{dv}{dx} &=&\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}
\frac{dy}{dx}=y^{2}+2xy\frac{dy}{dx}.
\end{eqnarray*}$$
To obtain $dy/dx$ combine the above results and solve the resulting equation for $dy/dx$. 
Remark: It is not necessary to introduce the functions $u$ and $v$. I have
introduce them to illustrate the general case, but you can do the
computation directly
$$\begin{equation*}
2\left( 2xy+x^{2}\frac{dy}{dx}\right) +3\left( y^{2}+2xy\frac{dy}{dx}\right)
=0.
\end{equation*}$$
Added: This method is a particular case of a general one. If you have an
implicit function $F(x,y)=0$ we can find $dy/dx$ by differentiating both sides of the implicit equation and solve for  $dy/dx$
$$\begin{eqnarray*}
\frac{dF}{dx} &=&\frac{\partial F}{\partial x}\frac{dx}{dx}+\frac{\partial F
}{\partial y}\frac{dy}{dx}=0 \\
&\Rightarrow &\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}
\frac{dy}{dx}=0 \\
&\Leftrightarrow &\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.
\end{eqnarray*}$$
