indeed, integration by parts it isdoes the trick:
$$\int d\mathbf{x}\; \vec{x} \,f(\vec{x})\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)=-\int d\mathbf{x}\; \,f(\vec{x})\,\mathbf{C}\cdot\frac{\partial}{\partial \vec{x}}\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)$$$$\int d\mathbf{x}\; \mathbf{x} \,f(\mathbf{x})\exp\left(-\tfrac{1}{2}\mathbf{x}\cdot \mathbf{C}^{-1}\cdot\mathbf{x}\right)=-\int d\mathbf{x}\; \,f(\mathbf{x})\,\mathbf{C}\cdot\frac{\partial}{\partial \mathbf{x}}\exp\left(-\tfrac{1}{2}\mathbf{x}\cdot \mathbf{C}^{-1}\cdot\mathbf{x}\right)$$ $$=\int d\mathbf{x}\; \,\exp\left(-\tfrac{1}{2}\vec{x}\cdot \mathbf{C}^{-1}\cdot\vec{x}\right)\mathbf{C}\cdot\frac{\partial}{\partial \vec{x}}\, f(\vec{x})$$
where$$=\int d\mathbf{x}\; \,\exp\left(-\tfrac{1}{2}\mathbf{x}\cdot \mathbf{C}^{-1}\cdot\mathbf{x}\right)\mathbf{C}\cdot\frac{\partial}{\partial \mathbf{x}}\, f(\mathbf{x})$$ $$\Rightarrow\left\langle\mathbf{x}\,f(\mathbf{x})\right\rangle=\mathbf{C}\cdot\left\langle \frac{\partial}{\partial \mathbf{x}}f(\mathbf{x})\right\rangle$$ where I have used that $\mathbf{C}$ is a symmetric matrix