Find all functions $f \colon \mathbb R \to \mathbb R$ such that for $\forall x, y \in \left[-\dfrac{1}{2}, +\infty\right)$, $$f(x)f(y) + f\left(\frac{1}{2} + \sqrt{xy(x + y) + \frac{1}{4}}\right) = f(xy) + f(x + y)$$
Let $P(x, y)$ be the assertion of $f(x)f(y) + f\left(\dfrac{1}{2} + \sqrt{xy(x + y) + \dfrac{1}{4}}\right) = f(xy) + f(x + y)$.
We have that for $P(0, 0)$ and $P(0, 1)$, it can be seen respectively that $\left\{ \begin{align} f(0)[f(0) - 2] &= -f(1)\\ f(0)[f(1) - 1] &= 0 \end{align} \right.$, which implies that $\left[ \begin{align} f(0) = f(1) = 0\\ f(0) = f(1) = 1 \end{align} \right.$.
And that's all I have for now, beneath are some other miscellaneous ideas that contributed nothing to my thoughts.
For $$P\left(\dfrac{xy}{2} - \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}, \dfrac{xy}{2} + \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right)$$, it could be obtained that $$f(x)f(y) = f\left(\dfrac{xy}{2} - \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right) \cdot f\left(\dfrac{xy}{2} + \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right)$$
For $$P\left(\frac{(m - 1) - \sqrt{m^2 - 6m + 1}}{2}, \frac{(m - 1) + \sqrt{m^2 - 6m + 1}}{2}\right)$$, we could suppose that $$f\left(\frac{(m - 1) - \sqrt{m^2 - 6m + 1}}{2}\right) \cdot f\left(\frac{(m - 1) + \sqrt{m^2 - 6m + 1}}{2}\right) = f(2(m - 1))$$
The same goes for $$P\left(\frac{(2m - 1) - \sqrt{4m^2 - 20m - 7}}{4}, \frac{(2m - 1) + \sqrt{4m^2 - 20m - 7}}{4}\right)$$, it happens to be that $$f\left(\frac{(2m - 1) - \sqrt{4m^2 - 20m - 7}}{4}\right) \cdot f\left(\frac{(2m - 1) + \sqrt{4m^2 - 20m - 7}}{4}\right) = f(2(2m - 1))$$
That's all for now, thanks for checking in, even if you don't have anything to add to this problem~