Answer
$$\frac{dy}{dx}=\frac{3xy^2+7y}{1-3x^2y-7x}$$
Work Step by Step
$$(3xy+7)^2=6y$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(3xy+7)^2=\frac{d}{dx}(6y)$$ $$2(3xy+7)\frac{d}{dx}(3xy+7)=6\frac{dy}{dx}$$ $$2(3xy+7)3\frac{d}{dx}(xy)=6\frac{dy}{dx}$$ $$6(3xy+7)\Big(y+x\frac{dy}{dx}\Big)=6\frac{dy}{dx}$$ $$(3xy+7)\Big(y+x\frac{dy}{dx}\Big)=\frac{dy}{dx}$$ $$(3xy^2+7y)+(3xy+7)x\frac{dy}{dx}=\frac{dy}{dx}$$ $$(3xy^2+7y)+(3x^2y+7x)\frac{dy}{dx}=\frac{dy}{dx}$$
2) Collect the terms with $dy/dx$ on one side and solve for $dy/dx$: $$(3x^2y+7x)\frac{dy}{dx}-\frac{dy}{dx}=-(3xy^2+7y)$$ $$(3x^2y+7x-1)\frac{dy}{dx}=-(3xy^2+7y)$$ $$\frac{dy}{dx}=-\frac{3xy^2+7y}{3x^2y+7x-1}=\frac{3xy^2+7y}{1-3x^2y-7x}$$
