Two functions of $x$ are differentiable and not identically zero. Find an example of two such functions having the property that the derivative of their quotient is the quotient of their derivatives.

1. We start with two differentiable functions \( f(x) \) and \( g(x) \) defined on an interval \( I \subseteq \mathbb{R} \), where \( g'(x) \neq 0 \) for all \( x \in I \). We need to find an example where the derivative of their quotient is equal to the quotient of their derivatives:
   \[
   \left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)}{g'(x)}
   \]

2. Using the quotient rule for differentiation, we have:
   \[
   \left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}
   \]

3. According to the problem, this should be equal to:
   \[
   \frac{f'(x)}{g'(x)}
   \]

4. Equating the two expressions, we get:
   \[
   \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} = \frac{f'(x)}{g'(x)}
   \]

5. Cross-multiplying to clear the denominators, we obtain:
   \[
   f'(x)g(x)g'(x) - f(x)[g'(x)]^2 = f'(x)g(x)^2
   \]

6. Rearranging terms, we get:
   \[
   f'(x)g(x)g'(x) - f'(x)g(x)^2 = f(x)[g'(x)]^2
   \]

7. Factoring out \( f'(x)g(x) \) on the left-hand side, we have:
   \[
   f'(x)g(x)(g'(x) - g(x)) = f(x)[g'(x)]^2
   \]

8. Dividing both sides by \( f(x)g(x) \), we get:
   \[
   \frac{f'(x)}{f(x)} \cdot \frac{g'(x)}{g(x)} - \left( \frac{g'(x)}{g(x)} \right)^2 = \frac{f'(x)}{f(x)}
   \]

9. Let \( u(x) = \ln(f(x)) \) and \( v(x) = \ln(g(x)) \). Then \( u'(x) = \frac{f'(x)}{f(x)} \) and \( v'(x) = \frac{g'(x)}{g(x)} \). Substituting these into the equation, we get:
   \[
   u'(x) v'(x) - [v'(x)]^2 = u'(x)
   \]

10. Solving for \( u'(x) \), we have:
    \[
    u'(x) = \frac{[v'(x)]^2}{v'(x) - 1}
    \]

11. To find specific functions \( f(x) \) and \( g(x) \), we choose a continuous function \( v'(x) \) such that \( v'(x) \neq 1 \) for all \( x \in I \). For simplicity, let \( v'(x) = 2 \). Then:
    \[
    v(x) = \int 2 \, dx = 2x + C
    \]
    Without loss of generality, let \( C = 0 \), so \( v(x) = 2x \).

12. Then \( g(x) = e^{v(x)} = e^{2x} \).

13. Next, we find \( u'(x) \):
    \[
    u'(x) = \frac{[v'(x)]^2}{v'(x) - 1} = \frac{2^2}{2 - 1} = 4
    \]

14. Integrating \( u'(x) \), we get:
    \[
    u(x) = \int 4 \, dx = 4x + D
    \]
    Again, let \( D = 0 \), so \( u(x) = 4x \).

15. Then \( f(x) = e^{u(x)} = e^{4x} \).

Thus, the functions \( f(x) = e^{4x} \) and \( g(x) = e^{2x} \) satisfy the given condition.

The final answer is \( \boxed{ f(x) = e^{4x} } \) and \( g(x) = e^{2x} \).