1. **State the problem:** Find all critical points of the function $$f(x,y) = xy + \frac{6}{x} + \frac{7}{y}.$$ 2. **Recall the definition of critical points:** Critical points occur where the gradient of $f$, i.e., the vector of partial derivatives, is zero or undefined. 3. **Compute partial derivatives:** - Partial derivative with respect to $x$: $$f_x = \frac{\partial}{\partial x} \left(xy + \frac{6}{x} + \frac{7}{y}\right) = y - \frac{6}{x^2}.$$ - Partial derivative with respect to $y$: $$f_y = \frac{\partial}{\partial y} \left(xy + \frac{6}{x} + \frac{7}{y}\right) = x - \frac{7}{y^2}.$$ 4. **Set partial derivatives equal to zero to find critical points:** $$f_x = y - \frac{6}{x^2} = 0 \implies y = \frac{6}{x^2}.$$ $$f_y = x - \frac{7}{y^2} = 0 \implies x = \frac{7}{y^2}.$$ 5. **Substitute $y$ from the first equation into the second:** $$x = \frac{7}{\left(\frac{6}{x^2}\right)^2} = \frac{7}{\frac{36}{x^4}} = \frac{7}{36} x^4.$$ 6. **Solve for $x$:** Multiply both sides by 36 to clear the denominator: $$36x = 7x^4.$$ Rewrite: $$7x^4 - 36x = 0.$$ Factor out $x$: $$x(7x^3 - 36) = 0.$$ 7. **Find roots:** - $x = 0$ is not in the domain because of division by $x$ in $f$. - Solve $7x^3 - 36 = 0$: $$7x^3 = 36 \implies x^3 = \frac{36}{7} \implies x = \sqrt[3]{\frac{36}{7}}.$$ 8. **Find corresponding $y$:** $$y = \frac{6}{x^2} = \frac{6}{\left(\sqrt[3]{\frac{36}{7}}\right)^2} = 6 \cdot \left(\frac{7}{36}\right)^{\frac{2}{3}}.$$ 9. **Summary of critical point:** $$\left(\sqrt[3]{\frac{36}{7}}, 6 \cdot \left(\frac{7}{36}\right)^{\frac{2}{3}}\right).$$ 10. **Check domain restrictions:** $x \neq 0$, $y \neq 0$, which holds here. **Final answer:** The function has one critical point at $$\boxed{\left(\sqrt[3]{\frac{36}{7}}, 6 \left(\frac{7}{36}\right)^{\frac{2}{3}}\right)}.$$