Partial Derivative with product & chain rule I cannot for the life of me work out the answer to this partial derivative. 
$$\frac{\delta}{\delta x}\left(\frac{x^2}{(x+y)^2(x+z)^2}\right) $$
My first thought was:


*

*Split into two equations:


$$\frac{\delta}{\delta x}\left(\frac{x}{(x+y)^2}\cdot \frac{x}{(x+z)^2}\right)$$


*Apply chain rule to each side which gives me:


$$\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}$$ and $$ \frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3} $$


*I then try to use the product rule, so:


$$ \left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+z)^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+y)^2}\right\rbrack $$
However I don't end up anywhere near an answer, let alone the right answer.
Apparently the answer is:
$$ -\frac{2(x^3-xyz)}{(x+y)^3(x+z)^3} $$
Is there an easier way to do this?

$$\left\lbrack\left(\frac{1}{(x+y)^2}-\frac{2x}{(x+y)^3}\right)\cdot \frac{x}{(x+{\bf{z}})^2}\right\rbrack + \left\lbrack\left(\frac{1}{(x+z)^2}-\frac{2x}{(x+z)^3}\right)\cdot \frac{x}{(x+{\bf{y}})^2}\right\rbrack = \\
\frac{x(y-x)}{(x+y)^3(x+z)^2}+\frac{x(z-x)}{(x+z)^3(x+y)^2} = \\
\frac{x(y-x)(x+z)+x(z-x)(x+y)}{(x+y)^3(x+z)^3} = \\
\frac{x^2y -x^3 +xyz-x^2z + x^2z -x^3 + xyz -x^2y}{(x+y)^3(x+z)^3} = \\
\frac{ -2x^3 +2xyz}{(x+y)^3(x+z)^3}. 
$$
