= Solution
The natural map between <Bott-Chern cohomology> and <Dolbeault cohomology> is
$$
\phi:[\alpha]_{BC}\longmapsto[\alpha]_{\bar\partial}.
$$
It is well-defined: a $d$-closed pure-type form is $\bar\partial$-closed, and $\partial\bar\partial\beta=-\bar\partial(\partial\beta)$ is $\bar\partial$-exact.
For surjectivity, choose the unique <harmonic differential form> $h$ representing a <Dolbeault cohomology> class using <Dolbeault Hodge decomposition>. The <Kähler Laplacian identity> $\Delta_d=2\Delta_{\bar\partial}$ makes $h$ a <harmonic differential form> for $d$, hence $d$-closed. Sending the Dolbeault class to $[h]_{BC}$ consequently gives a right inverse.
For injectivity, the <ddbar lemma> says directly that a $d$-closed, $\bar\partial$-exact pure-type form is $\partial\bar\partial$-exact. One can also exhibit its primitive. Let $G_d$ be the <Green operator of the Hodge Laplacian>, and put $G_{\bar\partial}=2G_d$. If $\alpha$ is $d$-closed and $\bar\partial$-exact, its harmonic projection vanishes and the <Kähler identities> give
$$
\alpha=\bar\partial\bar\partial^*G_{\bar\partial}\alpha
=-i\partial\bar\partial\Lambda G_{\bar\partial}\alpha.
$$
Here $\partial G_{\bar\partial}\alpha=\bar\partial G_{\bar\partial}\alpha=0$ because the Green operator commutes with these differentials, and $\bar\partial^*=-i[\Lambda,\partial]$. Thus $\alpha$ represents zero in <Bott-Chern cohomology>. We have constructed the canonical isomorphism and its harmonic inverse:
$$
\boxed{\phi:H_{BC}^{p,q}(M)\xrightarrow{\sim}H_{\bar\partial}^{p,q}(M),\qquad
\phi^{-1}([\alpha]_{\bar\partial})=[h]_{BC}.}
$$
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