The exercise reads as follows:
If $\mathcal X$ and $\mathcal Y$ are normed vector spaces, define $\alpha:\mathcal X^*\times\mathcal Y^*\to(\mathcal X\times\mathcal Y)^*$ by
$$\alpha(f,g)(x,y)=f(x)+g(y).$$
Then $\alpha$ is an isomorphism which is isometric if we use the norm $\|(x,y)\|=\max(\|x\|,\|y\|)$ on $\mathcal X\times\mathcal Y$, the corresponding operator norm on $(\mathcal X\times\mathcal Y)^*$, and the norm $\|(f,g)\|=\|f\|+\|g\|$ on $\mathcal X^*\times\mathcal Y^*$.
I have done everything but shown that $\alpha^{-1}$ is bounded and that $\alpha$ is an isometry.
For example, if $\alpha$ is an isometry, then
$$\|f+g\|_{(\mathcal X\times\mathcal Y)^*}=\|\alpha(f,g)\|_{(\mathcal X\times\mathcal Y)^*}=\|(f,g)\|_{\mathcal X^*\times\mathcal Y^*}=\|f\|_{\mathcal x^*}+\|g\|_{\mathcal Y^*},$$
which implies that $f$ and $g$ have the same sign everywhere, which makes no sense.
Observe that $$\|\alpha(f,g)\|=\sup_{\|(x,y)\|=1}|f(x)+g(y)|=\sup_{\|x\|\vee\|y\|=1}|f(x)+g(y)|\approx\|f\|+\|g\|=\|(f,g)\|,$$ where the approximation follows by taking $x,y$ with $\|x\|=\|y\|=1$ and $f(x)$ and $g(y)$ arbitrarily close to $\|f\|$ and $\|g\|$. It follows that $\alpha$ is an isometric isomorphism onto its image. It remains only to show that it is surjective. Indeed, if $h\in(X\times Y)^*$ then we can let $f(x)=h(x,0)$ and $g(y)=h(0,y)$ so that $f\in X^*$, $g\in Y^*$, and $h(x,y)=f(x)+g(y)$.