Definition. $X$ smooth vector field over a manifold $M$, the Lie derivative of $f\in C^{\infty}(M)$ along the flow of $X$ is: $\mathcal L_X f(m) = \langle df, X \rangle (m)$.
Definition. $M$ manifold, $f\in C^{\infty}(M)$ is a first integral of a smooth vector field $X$ if $\forall c\in f(M)$ the level set $f^{-1}(\{c\})$ is invariant wrt the flow $\phi_X$ of the vector field: $(\phi_X^t)^*f=f,\ \forall t$ (where $(\phi_X^t)^*f$ denotes the pullback of $f$ by $\phi_X^t$).
Theorem. $f\in C^{\infty}(M)$ is a first integral of $X$ iff $\mathcal L_X f=0$.
To prove the theorem professor gave us a suggestion: the proof follows by the definition of first integral, derivating along a curve and then evaluating at $t=0$.
Doing some computation $$\frac{d}{dt} (\phi_X^t)^*f\ |_{t=0} = \frac{d}{dt} (f\circ\phi_X^t)\ |_{t=0} = \langle df, X \rangle (\phi_X^t)\ |_{t=0} = (\phi_X^t)^*\mathcal L_Xf\ |_{t=0}$$
$\phi_X^0$ is the identity map, I guess we have to use this fact.
How to prove the two implications "$\Leftarrow$" and "$\Rightarrow$"?
I'll simplify notation and call $\phi_t \equiv \phi_X^t$, because there's only one vector field $X$ in question.
For $\implies$, by hypothesis we have that for all $t$, $\phi_t^*f = f$. This is constant in $t$, so in particular its derivative at $t=0$ has to vanish: \begin{align} 0 &= \dfrac{d}{dt} \bigg|_{t=0} \phi_t^*f \\ &= \phi_0^*(L_Xf) \tag{your calculation}\\ &= L_Xf. \end{align}
For the converse, we make the following observation. For any $t \in \Bbb{R}$, \begin{align} \dfrac{d}{dt} \bigg|_t \phi^*_tf &= \dfrac{d}{ds}\bigg|_{s=0}(\phi_{s+t})^*f \tag{chain rule} \\ &= \dfrac{d}{ds}\bigg|_{s=0}(\phi_s \circ \phi_t)^*f \tag{property of flow} \\ &= \dfrac{d}{ds}\bigg|_{s=0}\phi_t^*(\phi_s^*f) \tag{property of pullback} \\ &= \phi_t^* \left(\dfrac{d}{ds}\bigg|_{s=0} \phi_s^*f \right) \tag{$\ddot{\smile}$}\\ &= \phi_t^*(L_Xf) \tag{your calculation} \end{align} If $\ddot{\smile}$ isn't clear to you, write the derivative as a difference quotient, then use the fact that pullback respects all the usual algebraic operations.
Therefore, what this computation shows is that if $L_Xf = 0$, then for all $t$, the derivative $\dfrac{d}{dt}\bigg|_t \phi_t^*f = 0$ vanishes. If we assume that the range of values of $t$ is a connected interval in $\Bbb{R}$ (which is usually the case in the definition of flows I believe), then it follows that for all $t$, $\phi_t^*f = \phi_0^*f = f$ is constant in $t$.
This last part btw is just an application of the familiar single-variable arugment, whereby if a function $F : I \to \Bbb{R}$ ($0 \in I \subset$ being a connected interval) has vanishing derivative; i.e $F' = 0$, then $F$ is constant; in particular, for all $x \in I$, $F(x) = F(0)$.