Non-linear equation system Resolution I want to prove (I am not sure it is true though) that
$$\left\{\begin{array}{c} x-y-x\left( x^2+5y^2\right)=0\\ x+y-y\left( x^2+y^2\right) =0\end{array}\right.$$
Admits no other solution in the plane than the origin.
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I tried:
Let's consider $ \left( x,y\right)\in \mathbb{R}^2-\{  \left( 0,0\right)\} $
$y\neq0$ otherwise $x=0$
So I can change second equation as such:
$$ x^2+y^2= \frac{x+y}{y} $$
But that does not lead me anywhere actually.

As you have tried $x^2+y^2= \frac{x+y}{y}$, and then I tried, $x^2+5y^2=\frac{x-y}{x}$.. Now, from that I 've found $y^2=-\frac{x^2+y^2}{4xy}$. But $x^2+y^2>0$[assuming $\left( x,y\right)\in \mathbb{R}^2-\{  \left( 0,0\right)\}$]. So, $xy>0$. 
So, $x>0 \leftrightarrow y>0$ and $x<0 \leftrightarrow y<0$
Now, remark that $x^2+5y^2>x^2+y^2$. So, $\frac{x-y}{x}>\frac{x+y}{y}$. 
Now, we have two cases : $(1)x>0$ and $y>0$ (2)$x<0$ and $y<0$
Consider first case, 
$\frac{x-y}{x}>\frac{x+y}{y}$
$xy-y^2>xy+x^2$
$x^2+y^2<0$... which is not possible.
Now, consider second case, $\frac{x-y}{x}>\frac{x+y}{y}$
$[\frac{x-y}{x}]x<[\frac{x+y}{y}]x$ [because $x<0$]
$[\frac{x-y}{x}]xy>[\frac{x+y}{y}]xy$ [because $y<0$]
$y(x-y)>x(x+y)$
$xy-y^2>x^2+xy$
$x^2+y^2<0$ Which is not possible. So, We have $x=0$ and $y=0$
