The Euler's equation is $\rho \frac{d \vec{v}}{d t}=\vec{f}-\vec{\nabla} p=-\vec{\nabla}(p+\rho g z)$, where $z$ is vertically upwards.
Now $\frac{d \vec{v}}{d t}=\frac{\partial \vec{v}}{\partial t}+(\vec{v} \cdot \vec{\nabla}) \vec{v}$
$$
(\vec{v} \cdot \vec{\nabla}) \vec{v}=\vec{\nabla}\left(\frac{1}{2} v^{2}\right)-\vec{v} \times \operatorname{Curl} \vec{v}
$$
But 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) $\rho \vec{\nabla}\left(\frac{1}{2} v^{2}\right)=-\vec{\nabla}(p+\rho g z)$
or, $\vec{\nabla}\left(p+\frac{1}{2} \rho v^{2}+\rho g z\right)=0$
Hence
$$
p+\frac{1}{2} \rho v^{2}+\rho g z=\text { constant. }
$$