There is a hypothesis left over here: the flow is irrotational, i.e. the so-called vorticity
$$\vec{\omega}=\nabla\times\vec{v}$$
is zero. As a result, the velocity field is a gradient,
$$\vec{v}=\nabla\phi,$$
for some scalar field $\phi$: the field appearing in the Bernoulli equation you quoted. Note that the right-hand side is not necessarily zero but it can be taken as a constant, which does not depend either on position or on time.
This is a simple consequence of (i) Euler equation for an incompressible fluid,
$$\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\nabla) v + \nabla\left(\frac{p}{\rho}+\psi\right)=0,$$
where $\psi=gh$ but it could be the potential for any other conservative force field; and (ii) the identity
$$\vec{v}\times\vec{\omega} = \nabla\left(\frac{v^2}{2}\right) - (\vec{v}\cdot\nabla) v.$$
Then Euler equation reads
$$\nabla\left(\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi\right)=0.$$
Therefore the term between parentheses is function of time only but we can absorb any time dependence into $\phi$, and therefore we end up with
$$\frac{\partial\phi}{\partial t}+\frac{v^2}{2}+\frac{p}{\rho}+\psi=C$$
where $C$ is constant over space and time. This is therefore a stronger result than the traditional Bernoulli theorem where the constancy is only along each streamline, with a priori a different constant for each streamline. Here the constancy is over the entire fluid.
Best Answer
If you are studying something time independent then you just let $\frac{\partial}{\partial t}$ to be zero:
$$ \frac{\partial}{\partial x_j} \left [ \frac{1}{2} \rho v^2 v_j + \rho h v_j + \rho \phi v_j \right ]=0 $$
Next step is to get rid of $\rho$. Bernoulli's equation doesn't contain $\rho$, does it? We'll need mass balance for the stationary state:
$$\frac{\partial \rho v_j}{\partial x_j} = 0$$
which together with the first equation leads to: $$ \rho v_j \frac{\partial}{\partial x_j} \left [ \frac{1}{2} v^2 + h + \phi \right ]=0 $$
Now one should recall that Bernoulli's law is valid only along streamlines/pathlines, which do coincide for a steady flow. If something called $A$ is constant along the vector field $\boldsymbol v$, then it should suffice the equation
$$v_j \frac{\partial A}{\partial x_j} = 0$$
Actually it is just a derivative of $A$ along the vector field $\boldsymbol v$ --- if the derivative is zero then $A$ is constant along the vector field.
Thus we got the Bernoulli's law:
$$ \frac{1}{2} v^2 + h + \phi = const $$
along the streamline/pathline for the stationary case.