I have to prove the following question, but I'm having trouble. Any help would be greatly appreciated.
Let $V$ and $W$ be two (not necessarily finite dimensional) vector spaces. Show there is a natural isomorphism between the vector spaces $L(V,L(V,W))$ and $Bil(V \times V, W)$. Where $L(V,W)$ denotes the set of linear maps between $V$ and $W$ and $Bil(V \times V, W)$ denotes the space of bilinear maps $V \times V \rightarrow W$.
The keyword in the problem is "natural", so start by writing down a natural map between the two spaces, which is
$ \phi:\mathrm{Bil}(V\times V,W)\rightarrow L(V,L(V,W))\\ b\mapsto\;(\phi(b):V\rightarrow L(V,W),\; v\mapsto b(v,\cdot)). $
So $\phi(b)$ is the map that assigns to $v\in V$ the linear map $V\mapsto W, v^\prime\mapsto b(v,v^\prime)$, and as you already remarked, you have to show that $\phi$ is bijective and linear. To make sense of the last statement, you first have to define the vector space structure on $L(V,W)$ and thus on $L(V,L(V,W))$.
Surjectivity of $\phi$: let $F:V\rightarrow L(V,W)$ be a linear map. Then the map $b:V\times V\rightarrow W$ defined through the equation
$b(v_1,v_2):=F(v_1)(v_2)$
is bilinear and by definition $\phi(b)=F$.
Injectivity of $\phi$: $\phi(b)=0$ means that the linear map $b(v,\cdot)$ is the zero-map for every $v\in V$. That is
$\forall v\in V\;\forall v^\prime\in V\quad b(v,v^\prime)=0$,
which means that $b$ is the zero-map.