Problem: Let $f: \mathbb{R}^3 \to \mathbb{R}$ be a differentiable function such that $$y \frac{\partial f}{\partial x}(x,y,z) -x \frac{\partial f}{\partial y}(x,y,z) + \frac{\partial f}{\partial z}(x,y,z) \geq a>0, \forall(x,y,z) \in \mathbb{R}^3 \tag{*} $$ Let $\gamma: \mathbb{R}_+ \to \mathbb{R}^3$ a differentiable curve given by $\gamma(t)=(- \cos t, \sin t ,t), \ t \geq 0$
Show for $g(t):=f(\gamma(t))$ that $\lim_{t \to + \infty} g(t) = + \infty$
My approach: My idea was to make use of the fact that $g$ is differentiable and the derivative is given by $$g'(t)=< \nabla f(\gamma(t)), \dot\gamma(t)> $$ because the result will very much look like (*), if I can manage to show that $g'(t)\geq a >0$ it would follow that for all $t\geq 0$ I have that $g$ is strictly monotone increasing, if I then could prove that there exists no upper bound, this would finish the exercise.
I have $\dot \gamma(t)=(\sin t, \cos t ,1)$, my next step was to compute $g'(t)$, I obtained that $$g'(t)= \sin t \frac{\partial f( \gamma(t))}{\partial x}+ \cos t \frac{\partial f ( \gamma(t))}{\partial y} + \frac{\partial f(\gamma(t))}{\partial z} $$ Which looks a lot like (*), however not quite, because there is this annoying minus sign missing, which might be due to a typing error in C. Michels Analysis II Exercises.
Also I fail to make sense of the partial derivatives above, I left the $x,y,z$ in place in order to establish the connection with (*), however if I naively substitute $x= - \cos t, y = \sin t, z = t$ I get very weird partial derivatives that make no sense at all to me.
Is my approach to this exercise (in general) right? Or did I misguide myself into trap because (*) looks so much like the chain rule for curves.
Your approach is correct and you have done all the work. You got confused with a sign, but note that signs are exactly the ones you need.
Note that you have
$$g'(t)= \sin t \frac{\partial f( \gamma(t))}{\partial x}+ \cos t \frac{\partial f ( \gamma(t))}{\partial y} + \frac{\partial f(\gamma(t))}{\partial z}\ge a >0.$$
Why? Because $(x,y,z)=(-\cos t,\sin t,t).$ That is,
$$\sin t \frac{\partial f( \gamma(t))}{\partial x}+ \cos t \frac{\partial f ( \gamma(t))}{\partial y} + \frac{\partial f(\gamma(t))}{\partial z}$$
$$=y \frac{\partial f(x,y,z)}{\partial x}-x \frac{\partial f (x,y,z)}{\partial y} + \frac{\partial f(x,y,z)}{\partial z}.$$