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Here is a short geometric proof that if $R$ is Gorenstein and $\tilde{R}/R$ has dimension $\delta \le 3$, then $R$ is planar.
We can realize $R$ as the local ring of a rational curve $X$ of genus $\delta$. If $X$ is hyperelliptic (i.e. admit a degree $2$ morphism $f$ to $\mathbb{P}^{1}$), then $X$ embeds into a smoth surface: the ruled surface $\mathbb{P}(\mathcal{E})$ for $\mathcal{E}=f_{*}\mathcal{O}_{X}$. In particular, the singularities of $X$ are planar.
Otherwise, $X$ is non-hyperelliptic of genus $3$. But then the canonical map embeds $X$ as a plane quartic curve. In particular, $X$ again embeds in a smooth surface and hence has planar singularities.