You have $f(x,y)=(2x+3y,x+2y)$. This function is what we call linear, because it satisfies the following property: $f(\lambda v+w) = \lambda f(v)+f(w)$ for $v,w\in \Bbb R^2$ as you can check.
Now, linear functions are much easier do deal with than general functions. Why? Because they can be (at least in spaces like $\mathbb{R}^2$ that are what we call finite dimensional) be expressed in terms of matrices. Indeed, look that you have the following:
$$f(x,y)=xf(1,0)+yf(0,1)$$
Exactly because of that property of linearity. In that case, if you know about matrix multiplication, you'll see that this is equal to saying that
$$f(x,y)=\begin{pmatrix}f(1,0) & f(0,1)\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix},$$
where we consider elements of $\mathbb{R}^2$ as columns. In that case, you have
$$f(x,y)=\begin{pmatrix}2 & 3\\ 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix},$$
if we call that matrix $A$, then $f(v)=A v$ where $v=(x,y)$. Now, Look that because of that, things get easier. Indeed, to show that $f$ is injective, is equivalent to showing that $f(v)=f(w)$ implies $v=w$, but $f(v)=f(w)$ is the same as $Av=Aw$, or $A(v-w)=0$. This implies $v=w$, if and only if the homegenous system of linear equations $Av=0$ has only the trivial solution. That is, if and only if $\det A \neq 0$. But you can check that $\det A = 1$, so that $A$ is invertible, hence, $A(v-w)=0$ implies $v=w$, and hence $f$ is injective.
To show surjectivity, notice that $A$ being invertible, already implies that, because you can multiply the following
$$f(x,y)=A\begin{pmatrix}x\\y\end{pmatrix}$$
on the left by $A^{-1}$ to find $(x,y)$. It is easy, because
$$A^{-1}=\begin{pmatrix}-2 & \phantom{-}1 \\ \phantom{-}3 & -2\end{pmatrix},$$
and so we can take the inverse to be $g(v)=A^{-1}v$. In that case, $f(g(v))=A(A^{-1}v)=v$ and similarly $g(f(v))=A^{-1}(A(v))=v$, so $g$ is inverse to $f$ and hence $f$ is bijective.
