What is the gradient of $f (x, (y^2+z^2)^{1/2})$? Consider a $c \in \mathbb R$ and a function $f: U \subset \mathbb R^2\rightarrow \mathbb R $ with $ \nabla f (p) \neq 0, \forall p \in  f^{-1}(c)$, where $U$ is contained in the upper half plane y > 0.
Now define $g: U \times \mathbb R \rightarrow \mathbb R$ by $g(x, y, z) = f (x, (y^2+z^2)^{1/2})$. Then it is to be shown that $\nabla g(q) \neq 0, \forall q \in g^{-1}(c)$.
But I am not able to figure out how to write $\nabla f$ in way that would help me solve the problem.

I assume that what you want is to find $\nabla g$. In order to do so, we have to apply the chain rule. First of all, notice that the function $g$ can be interpreted as the composition of $f: U \subset \mathbb{R}^2\to\mathbb{R}$ and $h: \mathbb{R}^3\to\mathbb{R}^2$, where $h(x,y,z) = (x, (y^2+z^2)^\frac{1}{2})$. The composition is well defined and $g = f \:\circ h $.
Then the partial derivatives of $g$ are:

*

*$\frac{\partial g}{\partial x}(x,y,z) = f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2})$


*$\frac{\partial g}{\partial y}(x,y,z) = \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2})$


*$\frac{\partial g}{\partial z}(x,y,z) = \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2})$
The easiest way to compute the partial derivatives using the chain rule is, for me at least, using Jacobian matrices. More in particular:
$$
J_f(x,y) = \begin{bmatrix}
f^{(1,0)} & f^{(0,1)}  
\end{bmatrix}
$$
$$
J_h(x,y,z)= \begin{bmatrix}
1 & 0 & 0\\
0 & \frac{y}{(y^2+z^2)^\frac{1}{2}} & \frac{z}{(y^2+z^2)^\frac{1}{2}}
\end{bmatrix}
$$
The chain rule tell us that $J_g(x,y,z) = J_f(h(x,y,z))J_h(x,y,z)$. Therefore:
$$
J_g(x,y,z) = \begin{bmatrix}
f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2}) & \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}) & \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2})
\end{bmatrix}
$$
Therefore its gradient is:
$$\nabla g (x,y,z) = (f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2}), \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}), \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}))$$
