Show that $B(X,Y^*)$ and $B(Y,X^*)$ are isometrically isomorphic.

Given a bounded operator $T:X \to Y^{*}$ define $S: Y \to X^{*}$ by $(Sy)(x)=(Tx)(y)$. It is fairly routine to very that this is an isometric isomorphism. I will be glad to provide details if you get stuck.

Follow your nose. We want to define an isometry $\Phi:\mathcal{B}(X,Y^*)\to \mathcal{B}(Y,X^*)$, right? So take $T:X\to Y^*$. We want to define $\Phi(T):Y\to X^*$. This means I have to tell you what is the element $\Phi(T)(y)\in X^*$. In other words, I have to tell you what is the scalar $\Phi(T)(y)(x)$. But with the ingredients $T$, $y$ and $x$ we have, the only reasonable choice is $$\Phi(T)(y)(x)=T(x)(y).$$ Note that the right side is the element $T(x)\in Y^*$ applied in the element $y\in Y$, so this compiles. Since the expression $T(x)(y)$ is linear in $x$ (because $T$ is linear) and in $y$ (because $T(x)$ is a linear map too), and $T$ and $T(x)$ are continuous, this means that our definition works and the proposed codomain for $\Phi$ is correct.

The inverse of $\Phi$ is defined by the same construction, switching the roles of $X$ and $Y$.

So what's is left to do is checking that $\Phi$ is norm-preserving. We have $$|\Phi(T)(y)(x)|=|T(x)(y)|\leq \|T(x)\|\|y\|\leq \|T\|\|x\|\|y\|.$$This shows that $\|\Phi(T)\|\leq \|T\|$. The other inequality is similar.