6. use implicit differentiation to find ( y ) for ( y^{3}+x^{3}=x^{3}y^{3}+1 ).

Explanation:

Step1: Differentiate both sides of the equation

Differentiate (y^{3}+x^{3}=x^{3}y^{3}+1) with respect to (x). Using the sum rule ((u + v)^\prime=u^\prime+v^\prime), the power rule ((x^n)^\prime = nx^{n - 1}), and the product rule ((uv)^\prime=u^\prime v+uv^\prime) (where (u = x^{3}) and (v = y^{3})). For the left - hand side: (\frac{d}{dx}(y^{3}+x^{3})=\frac{d}{dx}(y^{3})+\frac{d}{dx}(x^{3})). By the chain rule (\frac{d}{dx}(y^{3}) = 3y^{2}y^\prime) and (\frac{d}{dx}(x^{3})=3x^{2}). For the right - hand side: (\frac{d}{dx}(x^{3}y^{3}+1)=\frac{d}{dx}(x^{3}y^{3})+\frac{d}{dx}(1)). Using the product rule, (\frac{d}{dx}(x^{3}y^{3})=3x^{2}y^{3}+x^{3}\times3y^{2}y^\prime) and (\frac{d}{dx}(1) = 0). So we have (3y^{2}y^\prime+3x^{2}=3x^{2}y^{3}+3x^{3}y^{2}y^\prime).

Step2: Solve for (y^\prime)

First, move all terms with (y^\prime) to one side: (3y^{2}y^\prime-3x^{3}y^{2}y^\prime=3x^{2}y^{3}-3x^{2}). Factor out (y^\prime) on the left - hand side: (y^\prime(3y^{2}-3x^{3}y^{2})=3x^{2}(y^{3}-1)). Then (y^\prime=\frac{3x^{2}(y^{3}-1)}{3y^{2}(1 - x^{3})}). Simplify the fraction: (y^\prime=\frac{x^{2}(y^{3}-1)}{y^{2}(1 - x^{3})}).

Answer:

(y^\prime=\frac{x^{2}(y^{3}-1)}{y^{2}(1 - x^{3})})