Functional equation related to $\sin$: $f(x+y)=f(x)f'(y)+f'(x)f(y)$

With $y=0$ you get $f(x)=f(x)f'(0)+f'(x)f(0)$.

If $f(0)\ne0$, we obtain $f'(x)=\frac{1}{f(0)}f(x)(1-f'(0))$. The case $f'(0)=1$ yields $f'(x)=0$, so $f$ is constant $c$, which implies $c=0$: a contradiction.

If $f'(0)\ne1$, we can write $f'(x)=kf(x)$, with $k\ne0$, so $f(x)=ae^{kx}$ (with $a\ne0$). The main relation now is $$ ae^{k(x+y)}=ae^{kx}ake^{ky}+ake^{kx}ae^{ky} $$ that implies $1=2ak$. This is a solution.

If $f(0)=0$, the equation becomes $f(x)=f(x)f'(0)$ or $(1-f'(0))f(x)=0$. If $f'(0)\ne 1$, we get the constant $0$ function.

If $f(0)=0$ and $f'(0)=1$, the business becomes interesting.

Since $f'(0)\ne1$, the function is not constant, so there is $y_0$ with $f(y_0)\ne0$. In particular, for all $x$, $$ f'(x)=\frac{1}{f(y_0)}(f(x+y_0)-f(x)f'(y_0)) $$ which shows $f'$ is differentiable (and also that $f$ is infinitely differentiable). Thus we can differentiate the main relation with respect to $x$ and $y$: \begin{align} f'(x+y)&=f'(x)f'(y)+f''(x)f(y) \\ f'(x+y)&=f(x)f''(y)+f'(x)f'(y) \end{align} that implies $f(x)f''(y)=f''(x)f(y)$. Therefore, for all $x$, $$ f''(x)=rf(x) $$ where $r=f''(y_0)/f(y_0)$.

This is an easy differential equation.