Determine the image of the map Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by \begin{equation*}f: \begin{pmatrix}x \\ y\end{pmatrix}\rightarrow \begin{pmatrix}u \\ v\end{pmatrix}=\begin{pmatrix}x(1-y) \\ xy\end{pmatrix}\end{equation*} 
I want to determine th eimage $f(\mathbb{R}^2)$. 
$$$$ 
We have that 
\begin{equation*}f\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix} \ \text{ und } \ f\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}\end{equation*} 
Does this mean that $\text{im}(f)=\left \{\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0\end{pmatrix}\right \}$ ? Or do we have to do something else in this case where we don't have a linear map? 

$f$ is not linear. Hence, you can't assume that the image of $f$ is a subspace of $\mathbb R^2$.
Hint: Consider
$$
f(x,y)=x\begin{pmatrix}1-y\\y\end{pmatrix}.
$$
You see that the image of $f$ contains all lines through $\begin{pmatrix}1-y\\y\end{pmatrix}$ for some $y\in\mathbb R$.
