= Solution
Write
$$
\mathbf u=\nabla\phi+\beta\nabla\alpha,
\qquad
\mathcal L=-\beta\alpha_t-\frac12\mathbf u\cdot\mathbf u.
$$
Variation of $\phi$ gives $\nabla\cdot\mathbf u=0$. Variation of $\alpha$ gives
$$
\beta_t+\nabla\cdot(\beta\mathbf u)=0,
$$
and variation of $\beta$ gives
$$
\alpha_t+\mathbf u\cdot\nabla\alpha=0.
$$
Using incompressibility, the middle equation becomes
$$
\beta_t+\mathbf u\cdot\nabla\beta=0.
$$
These are the three required Euler-Lagrange equations.
Let $D_t=\partial_t+\mathbf u\cdot\nabla$. Since $D_t\alpha=D_t\beta=0$, differentiating $u_i=\partial_i\phi+\beta\partial_i\alpha$ gives
$$
D_tu_i
=\partial_i(D_t\phi)-u_j\partial_i u_j
=\partial_i\left(D_t\phi-\frac12\mathbf u^2\right).
$$
Now
$$
D_t\phi
=\phi_t+\mathbf u\cdot\nabla\phi
=\phi_t+\mathbf u^2+\beta\alpha_t,
$$
because $\mathbf u\cdot\nabla\alpha=-\alpha_t$. Hence
$$
D_tu_i
=\partial_i\left(\phi_t+\beta\alpha_t+\frac12\mathbf u^2\right)
=-\partial_i p,
$$
where
$$
\boxed{p=-\frac12\mathbf u^2-\phi_t-\beta\alpha_t},
\qquad
\boxed{f(\phi_t,\alpha_t,\beta)=-\phi_t-\beta\alpha_t}.
$$
This is the <Clebsch-potential variational derivation of incompressible Euler flow>.
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