Question

y^{3}+x^{3}=9xy
\frac{dy}{dx}\frac{9y - 3x^{2}}{3y^{2}-9x}=\frac{3(3y - x^{2})}{3(y^{2}-3x)}=\frac{3y - x^{2}}{y^{2}-3x}

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(y^{3}+x^{3}=9xy\) term - by - term.
Using the chain rule \((y^{3})^\prime = 3y^{2}\frac{dy}{dx}\), \((x^{3})^\prime=3x^{2}\), and using the product rule \((9xy)^\prime = 9y + 9x\frac{dy}{dx}\).
We get \(3y^{2}\frac{dy}{dx}+3x^{2}=9y + 9x\frac{dy}{dx}\).

Step2: Rearrange the terms to solve for \(\frac{dy}{dx}\)

Move the terms with \(\frac{dy}{dx}\) to one side:
\(3y^{2}\frac{dy}{dx}-9x\frac{dy}{dx}=9y - 3x^{2}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(3y^{2}-9x)=9y - 3x^{2}\).
Then \(\frac{dy}{dx}=\frac{9y - 3x^{2}}{3y^{2}-9x}\).

Step3: Simplify the expression

Factor out 3 from the numerator and the denominator:
\(\frac{dy}{dx}=\frac{3(3y - x^{2})}{3(y^{2}-3x)}=\frac{3y - x^{2}}{y^{2}-3x}\).

Answer:

\(\frac{dy}{dx}=\frac{3y - x^{2}}{y^{2}-3x}\)