ar \( \frac{(x-y)^{2}}{2} \div \frac{2}{3(y-x)^{-2}}=3 / 4 \)

We start with the equation \[ \frac{(x-y)^2}{2} \div \frac{2}{3 (y-x)^{-2}} = \frac{3}{4}. \] **Step 1.** Rewrite the division as multiplication by the reciprocal: \[ \frac{(x-y)^2}{2} \times \frac{3 (y-x)^{-2}}{2} = \frac{3}{4}. \] **Step 2.** Notice that \((y-x)^{-2}\) means \[ (y-x)^{-2}=\frac{1}{(y-x)^2}. \] So, the expression becomes \[ \frac{(x-y)^2}{2} \times \frac{3}{2}\times \frac{1}{(y-x)^2} = \frac{3}{4}. \] **Step 3.** Recognize the relationship between \((x-y)^2\) and \((y-x)^2\). Since \[ (y-x) = -(x-y), \] raising to the power 2 gives \[ (y-x)^2 = (x-y)^2. \] **Step 4.** Substitute \((y-x)^2\) with \((x-y)^2\): \[ \frac{(x-y)^2}{2} \times \frac{3}{2}\times \frac{1}{(x-y)^2} = \frac{3}{4}. \] The \((x-y)^2\) cancels with \(\frac{1}{(x-y)^2}\) (provided \((x-y)^2\neq 0\), i.e. \(x\neq y\)): \[ \frac{3}{4} = \frac{3}{4}. \] **Step 5.** The identity \(\frac{3}{4}=\frac{3}{4}\) is always true. Therefore, the equation holds for any values of \(x\) and \(y\) provided that the cancellation was valid. The cancellation requires \[ (x-y)^2 \neq 0 \quad \Rightarrow \quad x\neq y. \] **Final Answer:** All pairs \((x, y)\) satisfying \(x\neq y\) are solutions.