Multiplying the given equality by $(a+b+c+d)$, we obtain
$$\begin{aligned} & \frac{a(a+b+c+d)}{b+c+d} + \frac{b(a+b+c+d)}{c+d+a} + \frac{c(a+b+c+d)}{d+a+b} + \frac{d(a+b+c+d)}{a+b+c} = \\ & = \frac{a^2+a(b+c+d)}{b+c+d} + \frac{b^2+b(c+d+a)}{c+d+a} + \frac{c^2+c(d+a+b)}{d+a+b} + \frac{d^2+d(a+b+c)}{a+b+c} = \\ & = \frac{a^2}{b+c+d} + a + \frac{b^2}{c+d+a} + b + \frac{c^2}{d+a+b} + c + \frac{d^2}{a+b+c} + d = a+b+c+d. \end{aligned}$$
It then follows that
$$\frac{a^2}{b+c+d} + \frac{b^2}{c+d+a} + \frac{c^2}{d+a+b} + \frac{d^2}{a+b+c} = 0.$$