Solution

The Euler's equation is $$\rho \dfrac{d\vec{v}}{dt}=\vec{f}-\vec{\nabla}p=-\vec{\nabla}(p+\rho\ gz)$$, where $$z$$ is vertically upwards.
Now $$\dfrac{d\vec{v}}{dt}=\dfrac{\partial\vec{v}}{\partial t}+(\vec{v}.\vec{nabla})\vec{v}$$.
But $$(\vec{v}.\vec{\nabla})\vec{v}=\vec{\nabla}\left(\dfrac{1}{2}v^{2}\right)-\vec{v}\times $$ curl $$\vec{v}$$
we consider the steady (i.e. $$\partial \vec{v}/\partial t=0$$) flow of an incompressible fluid then $$\rho=$$ constant. and as the motion is irrotational Curl $$\vec{v}=0$$
So from $$(1)$$ and $$(2)$$ $$\rho \vec{\nabla}\left(\dfrac{1}{2}v^{2}\right)=-\vec{nabla}(p+\rho gz)$$
or, $$\vec{\nabla}\left(p+\dfrac{1}{2}\rho v^{2}+\rho gz\right)=0$$
Hence $$p+\dfrac{1}{2}\rho v^{2}+\rho gz=$$ constant.