Let $G=Sl_2(\mathbb{F}_2)$ and $M= \mathbb{F}_2[x_1,x_2]$. 

$M$ is a $G$-module with $(A\cdot x_1, A\cdot x_2)=(x_1,x_2)\cdot A, (\forall) A \in Sl_2(\mathbb{F}_2)$.

I have to show that 

$M^G$=$\mathbb{F}_2[x_1\cdot v, x_1^2+ v]$, 

where $v=x_2\cdot(x_1+x_2)$ and $M^G=\lbrace m\in M| g\cdot m= m, \forall g\in G\rbrace$.

In general, how can I guess  the equality?

It's easy to prove ''$\supseteq$''.