How to solve this equation for $x$? How do I solve for $x$ from this equation?
$$ -\frac{1}{x^2} + \frac{9}{(4-x-y)^2} = 0.$$
I need to get this into $x=$"blah"?

If you add $\frac{1}{x^{2}}$ to both sides of your equation
$$-\frac{1}{x^{2}}+\frac{9}{\left( 4-x-y\right) ^{2}}=0\qquad (1)$$
you get this equivalent one (provided that $\frac{1}{x^{2}}$ is finite,
i.e $x\neq 0$)
$$\frac{1}{x^{2}}=\frac{9}{\left( 4-x-y\right) ^{2}}.\qquad (2)$$
It is satisfied if the square root of one side is equal or
symmetric to the square root of the other side: 
$$\frac{1}{x}=\pm \frac{3}{4-x-y}.\qquad (3)$$
Equation $(3)$ is equivalent to
$$3x=\pm \left( 4-x-y\right) \qquad (4)$$
provided that $x\neq 0$ and $4-x-y\neq 0$.
The equation $(4)$ represents two equations. One is 
$$3x=4-x-y,\qquad (5)$$
which is equivalent to
$$4x=4-y\Leftrightarrow x=1-\frac{1}{4}y\qquad (6)$$
and the other
$$3x=-\left( 4-x-y\right), \qquad (7)$$
is equivalent to
$$3x=-4+x+y\Leftrightarrow x=-2+\frac{1}{2}y.\qquad (9)$$
Thus $(1)$ is equivalent to
$$x=1-\frac{1}{4}y\qquad (10)$$
or
$$x=-2+\frac{1}{2}y,\qquad (11)$$
provided that $y\neq 4$ because the conditions $x\neq 0$ and $4-x-y\neq 0$
correspond to
$$x\neq 0\Leftrightarrow 1-\frac{1}{4}y\neq 0\Leftrightarrow y\neq 4$$
$$x\neq 0\Leftrightarrow -2+\frac{1}{2}y\neq 0\Leftrightarrow y\neq 4$$
and
$$4-x-y\neq 0\Leftrightarrow 4-\left( 1-\frac{1}{4}y\right) -y\neq
0\Leftrightarrow y\neq 4$$
$$4-x-y\neq 0\Leftrightarrow 4-\left( -2+\frac{1}{2}y\right) -y\neq
0\Leftrightarrow y\neq 4.$$
