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For computations you it is better to replace the first argument by a locally free resolution. This allows to compute the local $R{\mathcal H}om$. Then you can use the local-to-global spectral sequence to compute the global $RHom$. In your particular example, as everything happens in a neighborhood of $x$ and since $x$ is a locally complete intersection in $x$, you can find a vector bundle $E$ and it section in a neighborhood of $x$ such that its zero locus is $x$. Then the Koszul complex $$ 0 \to \Lambda^n E^* \to \dots \to \Lambda^2 E^* \to E^* \to O_X \to 0 $$ is a locally free resolution of the skyscraper. This shows that $$ {\mathcal E}xt^i(k(x),k(x)) = \Lambda^i E_x $$ and hence $$ Ext^i(k(x),k(x)) = \Lambda^i E_x $$ as well. Moreover, one has $E_x \cong T_xX$, so the final answer is $$ RHom(k(x),k(x)) = \oplus_{i=0}^n \Lambda^iT_xX[-i]. $$