= Solution
Vary the inverse <metric tensor> while holding the matter fields fixed. Differentiating $g_{a\mu}g^{\mu b}=\delta_a{}^b$ gives $\delta g_{ab}=-g_{a\mu}g_{b\nu}\delta g^{\mu\nu}$. The <Jacobi determinant derivative formula> consequently implies the <inverse metric volume variation>
$$
\delta\sqrt{-g}=-\frac12\sqrt{-g}\,g_{\alpha\beta}\delta g^{\alpha\beta}.
$$
Since the matter <Lagrangian> in this question contains no derivatives of the <metric tensor>, its metric variation is algebraic. The product rule gives
$$
\frac{\partial(\sqrt{-g}L)}{\partial g^{\alpha\beta}}
=\sqrt{-g}\left(\frac{\partial L}{\partial g^{\alpha\beta}}-\frac12g_{\alpha\beta}L\right).
$$
Multiplying by $-2/\sqrt{-g}$ proves \b[the equivalence of the stress-energy definitions]:
$$
\boxed{T_{\alpha\beta}=-2\frac{\partial L}{\partial g^{\alpha\beta}}+Lg_{\alpha\beta}.}
$$
The variations of the symmetric <metric tensor> are understood symmetrically. If the matter <action> contained metric derivatives, the definition would instead require a functional derivative and the corresponding integration by parts.
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