Square - Never ending problem!

${ x }^{ 2 }+\frac { { x }^{ 2 } }{ { (x+1) }^{ 2 } } =8$

$\Longleftrightarrow { x }^{ 2 }-\frac { 2{ x }^{ 2 } }{ (x+1) } +\frac { { x }^{ 2 } }{ { (x+1) }^{ 2 } } =8-\frac { 2{ x }^{ 2 } }{ (x+1) }$

$\Longleftrightarrow { (x-\frac { x }{ x+1 } })^{ 2 }=8-\frac { 2{ x }^{ 2 } }{ x+1 }$

$\Longleftrightarrow { (\frac { { x }^{ 2 } }{ x+1 } ) }^{ 2 }+\frac { 2{ x }^{ 2 } }{ x+1 } -8=0$

Let $t=\frac { { x }^{ 2 } }{ x+1 }$ . Because $x+1<0 \forall x<-1$ so $t<0$

Solve this equation: ${ t }^{ 2 }+2t-8=0$ .

We have 2 roots: ${t }_{ 1 }=2$ and ${ t }_{ 2 }=-4$

Because $t<0$ so we will choose ${ t }_{ 2 }=-4$

$\frac { { x }^{ 2 } }{ x+1 } =-4$ $\Longleftrightarrow x^{ 2 }+4x+4=0$ $\Longleftrightarrow x=-2$

So, $x=\boxed { -2 }$ is the answer!