Let the function satisfy $f(x)f'(-x)=f(-x)f'(x)$ and $f(0)=3$ for all $x$ Question
Let the function satisfy $f(x)f'(-x)=f(-x)f'(x)$ and $f(0)=3$ for all $x$.
Then find the number of roots of $f(x)=0$ in $[-2,2]$ and evaluate $\displaystyle\int\limits_{-51}^{51}\frac{\mathrm dx}{3+f(x)}$.

My Attempt
I have solved this equation:
\begin{align}
f(x)f'(-x)&=f(-x)f'(x);\\\\
\frac{f'(-x)}{f(-x)}&=\frac{f'(x)}{f(x)};\\\\
\int\frac{f'(-x)}{f(-x)}\,\mathrm dx&=\int\frac{f'(x)}{f(x)}\,\mathrm dx;\\\\
\int\frac{\mathrm df(-x)}{f(-x)}&=-\int\frac{\mathrm df(x)}{f(x)};\\\\
\log(f(-x))&=-\log(f(x))+\log(c).
\end{align}
Using the initial condition $f(0)=3$ I have found $$f(x)f(-x)=9,$$
but I do not know how to find $f(x)$ and hence how to find the number of roots of $f(x)$ in $[-2,2]$ and how to evaluate the given integral.
 A: Hint: You know that $f(x)f(-x)=9$. If $f(x)=0$, then what would be $f(-x)$?
For the integral, start by considering that $\int_{-a}^a g(x)dx=\int_{-a}^ag(-x)dx$.
A: OK, to conclude things faster note that
$$f(x)f'(-x)-f(-x)f'(x)=0 \tag{1}$$
Now, considering the product rule and chain rule together you can observe that
$$[f(x)f(-x)]'=0 \tag{2}$$
so by taking into account that $f(0)=3$ we can easily conclude that
$$f(x)f(-x)=9 \tag{3}$$
Next, consider that there is a $x_0 \in [-2,2]$ such that $f(x_0)=0$ but $x_0$ must also satisfy $(3)$ which leads to $0=9$ and hence a contradiction. So there is not such a $x_0$ and hence no roots!
Let us denote your integrals as
$$I=\int_{-a}^{a}\frac{1}{3+f(x)} dx\tag{4}$$
then use $(3)$ in conjunction with $(4)$ to conclude that
$$3I=\int_{-a}^{a}\frac{f(-x)}{3+f(-x)} dx\tag{5}$$
Now, use the general identity for definite integrals mentioned here with $(5)$ to get
$$3I=\int_{-a}^{a}\frac{f(x)}{3+f(x)} dx\tag{6}$$
Next, multiply $(4)$ by $3$ and add the result to $(6)$ to obtain
$$6I=\int_{-a}^{a} 1 dx = 2a$$
and finally
$$I=\color{blue}{\frac{a}{3}} \tag{7}$$
In your case, $a=51$.
