If $f(x+y)=f(x)f(y)-g(x)g(y)$ and $g(x+y)=f(x)g(y)+f(y)g(x)$, with $f'(0)=0$, determine $[f(x)]^2+[g(x)]^2$ Given the expressions:
$f(x+y)=f(x)f(y)-g(x)g(y)$
$g(x+y)=f(x)g(y)+f(y)g(x)$
the exercise is to show that $[f(x)]^2+[g(x)]^2$ is constant for all real $x$ and determine its value, knowing that $f$ and $g$ are real differentiable non-constant functions, and that $f'(0)=0$.
I realized it looks like the $\sin$ and $\cos$ functions, so the answer must be $1$. To prove something that way, I tried showing that $f$ and $g$ were always on the interval $[-1,1]$.
I have also tried to derivate each expression and plug in $x=y=0$ or only $y=0$, but was unable to develop the solution.

Let's derive your two equations, with respect to $x$ : you get
$$f'(x+y)=f'(x)f(y)-g'(x)g(y)$$
and
$$g'(x+y)=f'(x)g(y)+f(y)g'(x)$$
Evaluating them in $x=0$ gives you, because $f'(0)=0$, 
$$f'(y)=-g'(0)g(y)$$
and
$$g'(y)=f(y)g'(0)$$
So for all $y$,
$$f(y)f'(y) + g(y)g'(y) = - g'(0)g(y)f(y)+f(y)g'(0)g(y) = 0$$
You get that 
$$2(f(y)f'(y) + g(y)g'(y))=0, \quad \text{i.e. } (f^2+g^2)'(y)=0$$
So $f^2 + g^2$ is constant.
To find its value, plug $x=y=0$ in the two original equalities  : you get
$$f(0)=f^2(0)-g^2(0)$$
and
$$g(0)=2f(0)g(0)$$
If $g(0) \neq 0$, you would get from the second equation $2f(0)=1$, so $f(0)=\frac{1}{2}$. The first equation qives you $g^2(0)= \frac{1}{4}-\frac{1}{2} = -\frac{1}{2}$, which is impossible.
So $g(0)=0$, so the first equation gives you $f(0)=f^2(0)$, so $f(0)=0$ or $1$. But $f(0)=0$ is impossible because plugging $y=0$ in $f'(x+y)=f'(x)f(y)-g'(x)g(y)$ would say that $f'(x)=0$ and $f$ is constant, which is impossible. So $f(0)=1$, so you deduce that $f^2(0)+g^2(0)=1$.
Finally the value of the constant $f^2 + g^2$ is $1$.
