Let $M$ be a complex manifold with complex structure $J$ and Riemannian metric $g$. Then I know that $g$ is said "Hermitian" if it satisfies $g(X,Y)=g(JX,JY)$ for every $X,Y$ sections of the tangent bundle. Then it follows that $g$ is the real part of a sesquilinear form.
My question must be very stupid, but I'm not comfortable with this definition.. I know that $g$ is symmetric and that, considering $z^i$ complex coordinates and $z^j=x^j+iy^j$, it results $J(\frac{\partial}{\partial x^j})=\frac{\partial}{\partial y^j}$ and $J(\frac{\partial}{\partial y^j})=-\frac{\partial}{\partial x^j}$.
So if $g$ is Hermitian it follows $g(\frac{\partial}{\partial x^j},\frac{\partial}{\partial y^j})=g(J(\frac{\partial}{\partial x^j}),J(\frac{\partial}{\partial y^j}))=-g(\frac{\partial}{\partial y^j},\frac{\partial}{\partial x^j})=-g(\frac{\partial}{\partial x^j},\frac{\partial}{\partial y^j})$ i.e. $g(\frac{\partial}{\partial x^j},\frac{\partial}{\partial y^j})=0$
Am I right?
Yes, you're right. The point is that $J$ represents rotation by $90^\circ$ in the tangent space, and so $\dfrac{\partial}{\partial x^j}$ and $\dfrac{\partial}{\partial y^j} = J\left(\dfrac{\partial}{\partial x^j}\right)$ should, of course, be orthogonal.