If $\alpha, \beta$ are the roots of the equation $ x^2 - px + q = 0 $ and $\alpha_1, \beta_1$ are the roots of the equation $x^2 - qx + p = 0$,
Form the quadratic equation whose roots are $$ \frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} and \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta} $$
On
By Vieta's formulas: $$\alpha+\beta=p; \alpha\beta=q;\\ \alpha_1+\beta_1=q; \alpha_1\beta_1=p.$$ The equation to be found: $X^2+BX+C=0$, whose roots must be: $$\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} \ \ \text{and} \ \ \ \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}$$ The sum: $$-B=\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} + \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}=\frac{1}{\alpha_1}\cdot \frac{\alpha+\beta}{\alpha\beta}+\frac{1}{\beta_1}\cdot \frac{\alpha+\beta}{\alpha\beta}=\frac{\alpha+\beta}{\alpha\beta}\cdot \frac{\alpha_1+\beta_1}{\alpha_1\beta_1}=\cdots=1.$$ The product: $$C=\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{1}{\alpha_1^2\alpha\beta}+\frac{1}{\beta_1^2\alpha\beta}+\frac{1}{\alpha^2\alpha_1\beta_1}+\frac{1}{\beta^2\alpha_1\beta_1}=\\ \frac{1}{\alpha\beta}\cdot \frac{(\alpha_1+\beta_1)^2-2\alpha_1\beta_1}{\alpha_1^2\beta_1^2}+\frac{1}{\alpha_1\beta_1}\cdot \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha^2\beta^2}=\cdots=\\ \frac{q^3-2pq+p^3-2pq}{p^2q^2}.$$ Note: The triple dots to be filled by you.
Calculation gives $$\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)+\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{(\alpha +\beta)(\alpha_1+\beta_1)}{\alpha\alpha_1\beta\beta_1}=1$$ and similarly $$\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)*\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{p^3+q^3-4pq}{(pq)^2}$$ Then the equation is $$(pq)^2X^2-(pq)^2X+p^3+q^3-4pq=0$$