= Solution
For $x\in X$, define $e_x\in X^{**}$ by $e_x(\phi)=\phi(x)$. The map $J:x\mapsto e_x$ is linear, and
$$
|e_x(\phi)|\le\|\phi\|\,\|x\|
$$
shows $\|Jx\|\le\|x\|$, so $J$ is continuous. The granted equality $\|Jx\|=\|x\|$ is the <canonical embedding into the bidual>.
Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\|\cdot\|$ and $\|\cdot\|'$. Their common continuous dual has two operator norms, say $\|\cdot\|_*$ and $\|\cdot\|_*'$. Both dual spaces are <Banach space>[Banach], even if the second normed space $X$ is not complete. The identity map
$$
(X^*,\|\cdot\|_*)\longrightarrow(X^*,\|\cdot\|_*')
$$
has a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The <closed graph theorem> makes the two dual norms equivalent.
Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such that
$$
c\|x\|\le\|x\|'\le C\|x\|
\qquad(x\in X).
$$
This says the two norms are <equivalent norms>, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.
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