Question

x3+y3-3axy=0 find dy\dx

Answer 1

is it sufficient for you

Answer 2

${x}^{3} + {y}^{3} - 3axy = 0 \\ \\ \frac{d}{dx} \bigg( {x}^{3} + {y}^{3} - 3axy \bigg) = 0\\ \\ \frac{d}{dx} \big( {x}^{3} \big) + \frac{d}{dx} \big( {y}^{3} \big) - \frac{d}{dx} \big( 3axy \big) = 0 \\ \\ 3 {x}^{2} + 3 {y}^{2} \: \frac{dy}{dx} - 3a \bigg(y + x\frac{dy}{dx} \bigg) = 0 \\ \\ 3 {x}^{2} - 3ay + 3 {y}^{2} \: \frac{dy}{dx} - 3ax\: \frac{dy}{dx} = 0 \\ \\ 3 {x}^{2} - 3ay + \frac{dy}{dx} (3 {y}^{2} - 3ax)= 0 \\ \\ \frac{dy}{dx} (3 {y}^{2} - 3ax) = 3ay - 3 {x}^{2} \\ \\ \frac{dy}{dx} = \frac{3ay - 3 {x}^{2} }{3 {y}^{2} - 3ax} \\ \\ \boxed{ \red{\frac{dy}{dx} = \frac{ay - {x}^{2} }{ {y}^{2} - ax}}}$