1. Euler from Newton's Second Law

We derive the Euler momentum equation for an inviscid fluid by applying Newton's second law to a moving (material) fluid parcel and using the Reynolds transport theorem to relate integrals over a material volume to local fields. (The exam's wording mentions a "fixed control volume (fluid parcel)"; here we use a material volume $V(t)$ that moves with the fluid — i.e. the standard derivation.)

Let $\rho(\vec x,t)$ be the mass density and $\vec u(\vec x,t)$ the velocity field. Newton's second law for the material volume $V(t)$ (a fluid parcel) states

$$\frac{d}{dt}\int_{V(t)} \rho \vec u \, dV \;=\; \int_{V(t)} \rho \vec g \, dV + \int_{\partial V(t)} \vec t \, dS,$$

where $\vec g$ is the body force per unit mass (e.g. gravity) and $\vec t$ is the surface traction (force per unit area) acting on the boundary $\partial V(t)$.

For an inviscid fluid the only surface force is pressure, so $\vec t = -p\,\vec n$ where $\vec n$ is the outward unit normal on $\partial V$. Thus

$$\frac{d}{dt}\int_{V(t)} \rho \vec u \, dV = \int_{V(t)} \rho \vec g \, dV - \int_{\partial V(t)} p\,\vec n \, dS.$$

Apply the divergence theorem to the surface integral:

$$\int_{\partial V} p\,\vec n \, dS = \int_{V} \nabla p \, dV,$$

so

$$\frac{d}{dt}\int_{V(t)} \rho \vec u \, dV = \int_{V(t)} \rho \vec g \, dV - \int_{V(t)} \nabla p \, dV.$$

Now use the Reynolds transport theorem (RTT) for a material volume $V(t)$:

$$\frac{d}{dt}\int_{V(t)} \rho \vec u \, dV = \int_{V(t)} \left( \partial_{t}(\rho \vec u) + \nabla\cdot(\rho \vec u \otimes \vec u) \right)\,dV.$$

(Here $(\rho\vec u\otimes\vec u)_{ij}=\rho u_i u_j$.) Equating integrands (valid for arbitrary material volumes) gives the integral (and hence local) form:

$$\partial_{t}(\rho \vec u) + \nabla\cdot(\rho \vec u\otimes\vec u) = \rho \vec g - \nabla p.$$

This is the conservative form of the momentum equation. We can expand the left-hand side using the continuity equation

$$\partial_t{\rho}+ \nabla\cdot(\rho \vec u)=0.$$

Using product rule,

$$\partial_t(\rho\vec u) + \nabla\cdot(\rho\vec u\otimes\vec u) = \... ...\vec u\right) + \vec u\left(\partial_t{\rho} + \nabla\cdot(\rho\vec u)\right).$$

The term multiplying $\vec u$ vanishes by continuity, so we obtain the nonconservative (material-derivative) form:

$$\rho\left(\partial_t{\vec u} + (\vec u\cdot\nabla)\vec u\right) = -\nabla p + \rho \vec g.$$

Finally, for an incompressible fluid of constant density $\rho=$ const, divide by $\rho$ to obtain Euler's momentum equation in its familiar form

$$\boxed{\;\partial_t{\vec u} + (\vec u\cdot\nabla)\vec u = -\frac{1}{\rho}\nabla p + \vec g\; }$$

which is the desired result.