Establish natural isomorphism: $\mathcal{B}(E,F;G) \cong \mathcal{L}(E;\mathcal{L}(F;G))$ Where $\mathcal{B}(E,F;G)$ is the space of bilinear functions from vector spaces $E \times F \rightarrow G$ and $\mathcal{L}(E;\mathcal{L}(F;G))$ is the space of linear functions from $E \rightarrow \mathcal{L}(F;G)$. 
I don't know, in general, how to "establish an isomorphism." Please tell me if my solution is okay (for instance, if this were for a college course).

Let $\beta \in \mathcal{B}(E,F;G)$ and let $e\in E$. Let $\phi : \mathcal{B}(E,F;G) \rightarrow \mathcal{L}(E;\mathcal{L}(F;G))$ be defined as follows:
  $$(\phi(\beta))(e) = \beta (e, \cdot)$$
  Clearly $\beta (e, \cdot) : F \rightarrow G$ is linear, so $\phi$ takes $\beta$ to a linear function. This linear function takes $e\in E$ to $\beta (e, \cdot)$. We must check that $\phi$ is linear and bijective. Let $\beta_1 , \beta_2 \in \mathcal{B}(E,F;G)$ and let $c_1, c_2$ be scalars. Then $$ (\phi(c_1\beta_1 + c_2 \beta_2))(e) = (c_1\beta_1 + c_2 \beta_2)(e, \cdot)= c_1\beta_1(e, \cdot) + c_2\beta_2(e, \cdot) = c_1\phi(\beta_1)(e) + c_2\phi(\beta_2)(e)$$ (where the second equality holds because of the linearity of $\mathcal{B}(E,F;G)$), proving linearity. To prove injectivity, suppose $$\phi(\beta_1)=\phi(\beta_2).$$ Then for each $e \in E$ and $f \in F$, $$\beta_1 (e,f) = \beta_2 (e,f)$$ implying $\beta_1 = \beta_2$. To prove surjectivity, let $\omega \in \mathcal{L}(E;\mathcal{L}(F;G))$. For each $e$, $\omega (e) \in \mathcal{L}(F;G)$. For each $f$, $(\omega (e))(f) \in G$. By the linearity of $\mathcal{L}(E;\mathcal{L}(F;G))$, $\omega$ is linear in $E$ and by the linearity of $\mathcal{L}(F;G)$, $(\omega(e))$ is linear in $F$. So $(\omega (e))(f)$ defines a bilinear function, and we can write $$(\omega (e))(f) = \beta(e,f)$$ for some $\beta \in \mathcal{B}(E,F;G)$. Then $(\phi(\beta))(e) = \beta(e,\cdot) = \omega (e)(\cdot)$ and $$\phi(\beta) = \omega.$$So $\phi$ is a linear bijection, and thus an isomorphism. 

In general, is that how one proves an isomorphism? (linearity, invertibility)? Any need to explicitly define $\phi^{-1}$? Did I abuse notation in my proof anywhere? All advice is helpful.
 A: I think the second part of your proof is not quite as clean as you might like it to be. Notably you are essentially saying that since the expression $\omega(e)$ is linear in $e$ and  the expression $(\omega(e))(f)$ is linear in $f$ (what you actually say is more sloppy than this, and what I should really say is that the maps $e\in E\mapsto\omega(e)$ and $f\in F\mapsto\omega(e)(f)$ are linear, which they are by definition) it follows that the expression $(\omega(e))(f)$ is linear in both $e$ and $f$. That is essentially true (and essentially trivial), but note that the first part is not quite talking about the same expression as the second.
Rather than repair this, let me restate this slightly differently.
Define a map $\def\B{\mathcal{B}}\def\L{\mathcal{L}}\psi:\mathcal{L}(E;\L(F;G))\to\B(E,F;G)$ satisfying $\psi(\omega)=\bigl((e,f)\mapsto\omega(e)(f)\bigr)$. [Note that I dropped a pair of parentheses around $\omega(e)$, since $\omega(e)(f)$ is unambiguous.] To see that this is well defined, one must check that $\psi(\omega)$ is always bilinear. This is purely formal; here it goes. Fixing $f=f_0$ for linearity in the first argument: $$\psi(\omega)(\lambda e_1+\mu e_2,f_0) =\omega(\lambda e_1+\mu e_2)(f_0) =(\lambda\omega(e_1)+\mu\omega(e_2))(f_0) 
=\lambda\omega(e_1)(f_0)+\mu\omega(e_2)(f_0)\\
=\lambda\psi(\omega)(e_1,f_0)+\mu\psi(\omega)(e_2,f_0),
$$
where I've applied the definition of$~\psi$, then linearity of$~\omega$, then the definition of linear combinations of linear functions $F\to G$, and finally the definition of$~\psi$ again, in reverse. Similarly fixing $e=e_0$:
$$\psi(\omega)(e_0,\lambda f_1+\mu f_2) =\omega(e_0)(\lambda f_1+\mu f_2)  
=\lambda\omega(e_0)(f_1)+\mu\omega(e_0)(f_2)\\
=\lambda\psi(\omega)(e_0,f_1)+\mu\psi(\omega)(e_0,f_2),
$$
again using the definition of$~\psi$ twice, and in the middle linearity of $\omega(e_0)$.
With that done, and your $\phi$ which you showed well defined before considering injectivity, it remains to show that $\phi$ and $\psi$ are inverses (this removes the need to show that $\phi$ is injective or surjective; that follows immediately). Again this is purely formal (in fact even more so, only definitions are applied):
$$
  \psi(\phi(\beta)) = \bigl((e,f)\mapsto\phi(\beta)(e)(f)\bigr)
 =\bigl((e,f)\mapsto\beta(e,f)\bigr) = \beta
$$
and
$$
  \phi(\psi(\omega)) = \bigl(e\mapsto(f\mapsto \psi(\omega)(e,f))\bigr)
= \bigl(e\mapsto(f\mapsto \omega(e)(f))\bigr)
= \bigl(e\mapsto(\omega(e))\bigr) = \omega.
$$
