<s>[INST] Is it actually possible to solve $\frac{xy}{\ln(x^p)(x+y)^2}=c$ for $x$? Is it possible to solve the following for $x$
$$\frac{xy}{\ln(x^p)(x+y)^2}=c$$
where $p\in[0,1]$ and all other variables are positive real numbers ?

Without the $\ln(x^p)$ in the denominator, the solution simply is
$$x=-\frac{-y+2cy+y\sqrt{-4c+1}}{2c},\:x=-\frac{-y+2cy-y\sqrt{-4c+1}}{2c}$$
But does $\ln(x^p)$ make a solution impossible? 
Is it possible to solve it $p$ assumed values like $1,0.5$ etc.?
 [/INST] Probably, the best you can do (without numerical method) is to express $y$ in terms of $x$:
$$
\frac{(x+y)^2}{xy} = \frac{1}{c\ln(x^p)}
$$
$$
\frac{x}{y} + \frac{y}{x} + 2 = \frac{1}{c\ln(x^p)}
$$
Let $z=\frac{y}{x}$, then
$$
z+\frac{1}{z}=\frac{1}{c\ln(x^p)}-2
$$ 
$$
z=\frac{\frac{1}{c\ln(x^p)}-2\pm\sqrt{\frac{1}{c^2\ln^2 x^p}-\frac{4}{c\ln(x^p)}}}{2}
$$ 
$$
y=\frac{x(1-2c\ln(x^p)\pm\sqrt{1-4c\ln(x^p)}\;)}{2c\ln(x^p)}
$$
or
$$
y=\frac{x(1-2cp\ln(x)\pm\sqrt{1-4cp\ln(x)}\;)}{2cp\ln(x)}
$$
BTW, this means that $x\le e^{\frac{1}{4cp}}$.
</s>