Differentiate $11x^5 + x^4y + xy^5=18$

differentiating each term

$$\frac{d}{dx} (11x^5)=11(5x^4) = 55x^4$$

$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4\color{red}{\frac{dy}{dx}}] $$

$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4\color{red}{\frac{dy}{dx}}] $$

For each term involving both $x$ and $y$, differentiate the $x$ and $y$ parts separately and combine the two using the product rule: $$(x^4y)'=x^4(y)'+(x^4)'y=x^4\frac{dy}{dx}+4x^3y$$ $$(xy^5)'=x(y^5)'+(x)'y^5=5xy^4\frac{dy}{dx}+y^5$$ Thus $$55x^4+x^4\frac{dy}{dx}+4x^3y+5xy^4\frac{dy}{dx}+y^5=0$$ $$(x^4+5xy^4)\frac{dy}{dx}=-(55x^4+4x^3y+y^5)$$ $$\frac{dy}{dx}=-\frac{55x^4+4x^3y+y^5}{x^4+5xy^4}$$