Answer
The possible solutions are:
$$f(x)=10e^{x}~~\text{or}~~f(x)=-10e^{x}$$
Work Step by Step
Differentiate both sides with respect to $x$:
$$2f(x)f'(x)=\frac{d}{dx}\left(\int_{0}^{x}[(f(t))^{2}+(f'(t))^{2}]dt\right)$$
Using the $FTC$ it follows:
$$2f(x)f'(x)=(f(x))^{2}+(f'(x))^{2}$$
$$(f(x))^{2}-2f(x)f'(x)+((f'(x))^{2}=0$$
$$(f(x)-f'(x))^{2}=0$$
$$f'(x)=f(x)$$
$$\frac{f'(x)}{f(x)}=1$$
$$\int \frac{f'(x)}{f(x)}dx=\int1 dx$$
$$\ln(|f(x)|)=x+c$$
$$|f(x)|=e^{x+c}$$
$$f(x)=e^{x+c}~~\text{or}~~f(x)=-e^{x+c}$$
From the given equation when $x=0$ it follows:
$$(f(0))^{2}=100+0=100$$
$$f(0)=10~~\text{or}~~f(0)=-10$$
when $f(x) \gt 0$ it follows:
$$f(0)=e^{0+c}=10$$
This implies that $c=\ln(10)$
when $f(x) \lt 0$ it follows:
$$f(0)=-e^{0+c}=-10$$
This implies that $c=\ln(10)$
Therefore, the possible solutions are:
$$f(x)=e^{x+\ln(10)}~~\text{or}~~f(x)=-e^{x+\ln(10)}$$
$$f(x)=10e^{x}~~\text{or}~~f(x)=-10e^{x}$$
