how did he conclude that?integral So the question is : Find all continuous functions such that $\displaystyle \int_{0}^{x} f(t) \, dt= ((f(x)^2)+C$. 
Now in the solution, it starts with this, clearly $f^2$ is differentiable at every point ( its derivative is $f$). So $f(x)\ne0$? I have no idea how he concluded that, this is from Spivak's calculus, if you differentiate it's clearly $f(x)=f(x)f'(x)$ but he said that before even giving this formula help please?
EDIT : I know that $f(x)=f(x)f'(x)$ What  i don't understand is this "clearly $f^2$ is differentiable at every point ( its derivative is $f$). So $f(x)\ne0$" why $f(x)$ mustn't equal $0$?
 A: As DonAntonio pointed out, the left hand side is a constant (not a function like the right hand side). I'll assume that you made a typo and that you instead meant to type:
$$\int_0^x f(t)dt$$

Note that the left hand side is differentiable at every point, since by the fundamental theorem of calculus, we know that:
$$
\dfrac{d}{dx} \int_0^x f(t)dt = f(x)
$$
Hence, since the right hand side equals the left hand side, it follows that $(f(x))^2+C$ is also differentiable at every point.
A: I interpret the problem as follows: 
Find all continuous functions $f:\ \Omega\to{\mathbb R}$ defined in some open interval $\Omega$ containing the origin and satisfying the integral equation
$$\int_0^x f(t)\ dt=f^2(x)+ C\qquad(x\in\Omega)$$
for a suitable constant $C$.
Assume $f$ is such a function and that $f(x)\ne 0$ for some $x\in\Omega$. Then for $0<|h|\ll1$ we have
$$\int_x^{x+h} f(t)\ dt=f^2(x+h)-f^2(x)=\bigl(f(x+h)+f(x)\bigr)\bigl(f(x+h)-f(x)\bigr)\ ,$$
and using the mean value theorem for integrals we conclude that
$${f(x+h)-f(x)\over h}={f(x+\tau h)\over f(x+h)+f(x)}$$
for some $\tau\in[0,1]$. Letting $h\to0$ we see that $f$ is differentiable at $x$ and that $f'(x)={1\over2}$. It follows that the graph of $f$ is a line with slope ${1\over2}$ in all open intervals where $f$ is nonzero. When such a line coming from west-south-west arrives at the point $(a,0)$ on the $x$-axis then $f(a)=0$ by continuity, and similarly, when such a line starts at the point $(b,0)$ due east-north-east, then $f(b)=0$.
The above analysis leaves only the following possibility for such an $f$: There are constants $a$, $b$ with $-\infty\leq a\leq b\leq\infty$ such that
$$f(x)=\cases{-{1\over2}(a-x)\quad &$(x\leq a)$ \cr 0&$(a\leq x\leq b)$ \cr {1\over2}(x-b)&$(x\geq b)\ .$\cr}$$
It turns out that these $f$'s  are in fact solutions to the original problem. 
Proof. Assume for the moment
$$a\leq0\leq b\ .\tag{1}$$
When $0\leq x\leq b$ then $$\int_0^x f(t)\ dt=0=f^2(x)+0\ ,$$
and when $x\geq b$ then
$$\int_0^x f(t)\ dt=\int_b^x f(t)\ dt={1\over4}(t-b)^2\biggr|_b^x={1\over4}(x-b)^2= f^2(x)+0$$
as well. Similarly one argues for $x\leq0$.
In order to get rid of the assumption $(1)$ we note that when $f$ is a solution of the original problem then  any translate $g(x):=f(x-c)$, $\>c\in{\mathbb R}$, is a solution as well: For all $x\in{\mathbb R}$ we have
$$\eqalign{\int_0^x g(t)\ dt&=\int_0^x f(t-c)\ dt=\int_{-c}^{x-c} f(t')\ dt'=\int_{-c}^0 f(t')\ dt'+\int_0^{x-c} f(t')\ dt'\cr
&=\int_{-c}^0 f(t')\ dt'+f^2(x-c)+C=g^2(x)+C'\ .\cr}$$
