Find continuous functions satisfying given conditions

Find continuous functions which satisfy the given conditions for all $x \in \mathbb{R}$ .

No such function can exist since for $x = 0$ we have
$\int_0^x f(t) \, dt = \int_0^0 f(t) \, dt = 0 \neq e^0 = 1.$
Taking derivatives of both sides of the given equation we have
$\begin{align*} &&\frac{d}{dx} \left(\int_0^{x^2} f(t) \, dt\right) &= \frac{d}{dx} \left( 1 - 2^{(x^2)} \right) \\[9pt] \implies && (2x)(f(x^2)) &= \frac{d}{dx} \left(1 - e^{x^2 \log 2} \right) \\[9pt] \implies && (2x)(f(x^2)) &= -(2x \log 2) e^{x^2 \log 2} \\[9pt] \implies && f(x^2) &= -(\log 2)(2^{(x^2)}) \\[9pt] \implies && f(x) &= - 2^x \log 2. \end{align*}$
Again, taking derivatives of both sides we have
$\begin{align*} &&\frac{d}{dx} \left( \int_0^x f(t) \, dt \right) &= \frac{d}{dx} \left( f^2(x) - 1 \right) \\[9pt] \impies && f(x) &= 2f(x)f'(x) \\[9pt] \implies && 1 &= 2f'(x) \end{align*}$

at all points $x$ such that $f(x) \neq 0$ . (Since $\int_0^x f(t) \, dt = f^2(x) - 1$ is not satisfied by the zero function $f(x) = 0$ , we know there are real $x$ such that $f(x) \neq 0$ .) Then, integrating

$f'(x) = \frac{1}{2} \quad \implies \quad f(x) = \frac{1}{2} x + C.$

Now, we can solve for $C$ by evaluating the given identity at $x = 0$ ,

$\begin{align*} && \int_0^x f(t) \, dt &= f^2 (x) - 1 \\[9pt] \implies && \int_0^0 f(t) \, dt &= f^2 (0) - 1 \\[9pt] \implies && 0 &= \left( \frac{0}{2} + C\right)^2 - 1\\[9pt] \implies && 0 &= C^2 - 1 \\[9pt] \implies && C &= \pm 1. \end{align*}$

Therefore, we have

$f(x) = \frac{1}{2} x \pm 1.$