$ g(x)$ is a differentiable function for $ 0\leq x\leq \pi$ and $ g'(x)$ is a continuous function for $ 0\leq x\leq \pi$.
Let $ f(x) \equal{} g(x)\sin x.$ Find $ g(x)$ such that $ \int_0^{\pi} \{f(x)\}^2dx \equal{} \int_0^{\pi}\{f'(x)\}^2dx$.
1. Given the function \( f(x) = g(x) \sin x \), we need to find \( g(x) \) such that 
   \[
   \int_0^{\pi} \{f(x)\}^2 \, dx = \int_0^{\pi} \{f'(x)\}^2 \, dx.
   \]

2. First, we compute \( \int_0^{\pi} \{f(x)\}^2 \, dx \):
   \[
   \int_0^{\pi} \{f(x)\}^2 \, dx = \int_0^{\pi} \{g(x) \sin x\}^2 \, dx = \int_0^{\pi} g^2(x) \sin^2 x \, dx.
   \]
   Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \), we get:
   \[
   \int_0^{\pi} g^2(x) \sin^2 x \, dx = \int_0^{\pi} g^2(x) \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^{\pi} g^2(x) \, dx - \frac{1}{2} \int_0^{\pi} g^2(x) \cos(2x) \, dx.
   \]
   Since \( g(x) \cos(2x) \) is an odd function over the interval \( [0, \pi] \), the integral of an odd function over a symmetric interval is zero:
   \[
   \int_0^{\pi} g^2(x) \cos(2x) \, dx = 0.
   \]
   Therefore,
   \[
   \int_0^{\pi} g^2(x) \sin^2 x \, dx = \frac{1}{2} \int_0^{\pi} g^2(x) \, dx.
   \]

3. Next, we compute \( \int_0^{\pi} \{f'(x)\}^2 \, dx \):
   \[
   f'(x) = \frac{d}{dx} [g(x) \sin x] = g'(x) \sin x + g(x) \cos x.
   \]
   Therefore,
   \[
   \{f'(x)\}^2 = (g'(x) \sin x + g(x) \cos x)^2 = g'(x)^2 \sin^2 x + 2g'(x)g(x) \sin x \cos x + g(x)^2 \cos^2 x.
   \]
   Integrating this expression over \( [0, \pi] \):
   \[
   \int_0^{\pi} \{f'(x)\}^2 \, dx = \int_0^{\pi} \left( g'(x)^2 \sin^2 x + 2g'(x)g(x) \sin x \cos x + g(x)^2 \cos^2 x \right) \, dx.
   \]
   We can split this into three integrals:
   \[
   \int_0^{\pi} g'(x)^2 \sin^2 x \, dx + \int_0^{\pi} 2g'(x)g(x) \sin x \cos x \, dx + \int_0^{\pi} g(x)^2 \cos^2 x \, dx.
   \]
   Using the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \), we get:
   \[
   \int_0^{\pi} g(x)^2 \cos^2 x \, dx = \int_0^{\pi} g(x)^2 \frac{1 + \cos(2x)}{2} \, dx = \frac{1}{2} \int_0^{\pi} g(x)^2 \, dx + \frac{1}{2} \int_0^{\pi} g(x)^2 \cos(2x) \, dx.
   \]
   Again, since \( g(x) \cos(2x) \) is an odd function over the interval \( [0, \pi] \), the integral of an odd function over a symmetric interval is zero:
   \[
   \int_0^{\pi} g(x)^2 \cos(2x) \, dx = 0.
   \]
   Therefore,
   \[
   \int_0^{\pi} g(x)^2 \cos^2 x \, dx = \frac{1}{2} \int_0^{\pi} g(x)^2 \, dx.
   \]
   The middle term \( \int_0^{\pi} 2g'(x)g(x) \sin x \cos x \, dx \) is zero because it is the integral of an odd function over a symmetric interval:
   \[
   \int_0^{\pi} 2g'(x)g(x) \sin x \cos x \, dx = 0.
   \]
   Therefore,
   \[
   \int_0^{\pi} \{f'(x)\}^2 \, dx = \int_0^{\pi} g'(x)^2 \sin^2 x \, dx + \frac{1}{2} \int_0^{\pi} g(x)^2 \, dx.
   \]

4. Equating the two integrals:
   \[
   \frac{1}{2} \int_0^{\pi} g^2(x) \, dx = \int_0^{\pi} g'(x)^2 \sin^2 x \, dx + \frac{1}{2} \int_0^{\pi} g(x)^2 \, dx.
   \]
   Subtracting \( \frac{1}{2} \int_0^{\pi} g(x)^2 \, dx \) from both sides:
   \[
   0 = \int_0^{\pi} g'(x)^2 \sin^2 x \, dx.
   \]
   Since \( \sin^2 x \) is positive for \( 0 < x < \pi \), the only way this integral can be zero is if \( g'(x) = 0 \) for all \( x \) in \( [0, \pi] \). Therefore, \( g(x) \) must be a constant function.

5. Let \( g(x) = c \), where \( c \) is a constant.

The final answer is \( \boxed{ g(x) = c } \), where \( c \) is a constant.