We are given the equation:

$x - iy = \sqrt{\frac{a - ib}{c - id}}$

To start the proof, we first square both sides of the equation to remove the square root. This yields the relation:

$(x - iy)^2 = \frac{a - ib}{c - id}$

To find the term $(x + iy)^2$, we take the conjugate of the equation derived above. The conjugate of $(x - iy)^2$ is $(x + iy)^2$, and the conjugate of the fraction $\frac{a - ib}{c - id}$ is $\frac{a + ib}{c + id}$. This gives us the second equation:

$(x + iy)^2 = \frac{a + ib}{c + id}$

From the properties of complex numbers, we know that $x^2 + y^2 = (x + iy)(x - iy)$. Therefore, the square of this expression is:

$(x^2 + y^2)^2 = (x + iy)^2 (x - iy)^2$

We substitute the values from our derived equations into this identity.

Multiplying the two expressions, we get:

$(x^2 + y^2)^2 = \left( \frac{a + ib}{c + id} \right) \left( \frac{a - ib}{c - id} \right)$

Grouping the terms, this simplifies to:

$\frac{(a + ib)(a - ib)}{(c + id)(c - id)}$

Finally, using the identity $(A + iB)(A - iB) = A^2 + B^2$, we simplify the numerator to $a^2 + b^2$ and the denominator to $c^2 + d^2$. Hence, we have proved that:

$\left( x^2 + y^2 \right)^2 = \frac{a^2 + b^2}{c^2 + d^2}$