Preve that \[ \frac{(x-y)^{2}}{2} \div \frac{2}{3(y-x)^{-2}}=3 / 4 \]

Let's start with the left side of the equation: \[ \frac{(x-y)^{2}}{2} \div \frac{2}{3(y-x)^{-2}} \] First, we can rewrite the division as multiplication by the reciprocal: \[ \frac{(x-y)^{2}}{2} \times \frac{3(y-x)^{2}}{2} \] Now notice that \(y - x = - (x - y)\). Therefore, \((y - x)^{2} = (-(x - y))^{2} = (x - y)^{2}\). Thus, we can simplify: \[ = \frac{(x-y)^{2}}{2} \times \frac{3(x-y)^{2}}{2} \] Next, we multiply the fractions together: \[ = \frac{3(x-y)^{2} \cdot (x-y)^{2}}{4} \] This simplifies to: \[ = \frac{3(x-y)^{4}}{4} \] Now we need to check if this equals \(\frac{3}{4}\). For this equality to hold, \(3(x-y)^{4} = 3\), which implies: \[ (x-y)^{4} = 1 \] Thus: \[ x-y = 1 \quad \text{or} \quad x-y = -1 \] Therefore, the original equation does not hold true in general unless \(x-y = 1\) or \(x-y = -1\). Hence, the original expression simplifies to: \[ \frac{(x-y)^{2}}{2} \div \frac{2}{3(y-x)^{-2}} = \frac{3}{4} \] is indeed valid under specific conditions.