Determine all pairs of real numbers $(x, y)$ for which $\left(4 x^{2}-y^{2}\right)^{2}+(7 x+3 y-39)^{2}=0$.
First, we note that $a^{2} \geq 0$ for every real number $a$.

Therefore, $\left(4 x^{2}-y^{2}\right)^{2} \geq 0$ and $(7 x+3 y-39)^{2} \geq 0$.

Since $\left(4 x^{2}-y^{2}\right)^{2}+(7 x+3 y-39)^{2}=0$, then it must be the case that $\left(4 x^{2}-y^{2}\right)^{2}=0$ and $(7 x+3 y-39)^{2}=0$.

This means that $4 x^{2}-y^{2}=0$ and $7 x+3 y-39=0$.

The equation $4 x^{2}-y^{2}=0$ is equivalent to $4 x^{2}=y^{2}$ and to $y= \pm 2 x$.

If $y=2 x$, then the equation $7 x+3 y-39=0$ becomes $7 x+6 x=39$ or $x=3$.

This gives $y=2 x=6$.

If $y=-2 x$, then the equation $7 x+3 y-39=0$ becomes $7 x-6 x=39$ or $x=39$.

This gives $y=-2 x=-78$.

Therefore, the solutions to the original equation are $(x, y)=(3,6),(39,-78)$.

ANSWER: $(x, y)=(3,6),(39,-78)$