= Solution
Let $\mathcal L=-\tfrac12g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi-V$. In the <metric variation> specified here, the covariant <metric tensor> $g_{\mu\nu}$ is varied while the <scalar field> is held fixed. Differentiating the inverse and the determinant gives
$$
\delta g^{\alpha\beta}=-g^{\alpha\mu}g^{\beta\nu}\delta g_{\mu\nu},\qquad
\delta\sqrt{-g}=\frac12\sqrt{-g}\,g^{\mu\nu}\delta g_{\mu\nu}.
$$
Thus $\delta\mathcal L=\tfrac12\partial^\mu\phi\partial^\nu\phi\,\delta g_{\mu\nu}$, and the <action> varies as
$$
\delta S_M=\frac12\int d^4x\sqrt{-g}
\left(\partial^\mu\phi\partial^\nu\phi+g^{\mu\nu}\mathcal L\right)\delta g_{\mu\nu}.
$$
The functional derivative therefore gives $T^{\mu\nu}=\partial^\mu\phi\partial^\nu\phi+g^{\mu\nu}\mathcal L$. Lowering both indices yields the <stress-energy tensor of a canonical scalar field>:
$$
\boxed{T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi
-g_{\mu\nu}\left(\frac12g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi+V\right).}
$$
The determinant variation has a positive sign because the variable is $g_{\mu\nu}$; varying the inverse <metric tensor> instead would reverse that intermediate sign.
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