Implicitly differentiate $x^3y+3xy^3 =x+y$

Question

This is an online question, and the system is marking it to be incorrect. However, I can not figure out where I went wrong.

Calculate the derivative of $y$ with respect to $x$

$x^3y+3xy^3 =x+y$

Here is my attempt:

$(x^3y)'+(3xy^3)'=x'+y'$

$(3x^2y+\frac{dy}{dx}x^3)+3(y^3+3y^2\frac{dy}{dx}x)=1+\frac{dy}{dx}$

$3x^2y+\frac{dy}{dx}x^3+3y^3+9y^2\frac{dy}{dx}x=1+\frac{dy}{dx}$

$3x^2y+3y^3-1=\frac{dy}{dx}-\frac{dy}{dx}x^3-9y^2\frac{dy}{dx}x$

$3x^2y+3y^3-1=\frac{dy}{dx}(1-x^3-9y^2x)$

$\frac{3x^2y+3y^3-1}{1-x^3-9y^2x}=\frac{dy}{dx}$

Answer 1

Your work is correct, but when you typed in the answer, you wrote $$3x^3y$$ when you intended to write $$3x^2 y.$$

Answer 2

I just look through it, the reasoning is correct. However, when you typed in the answer you made a mistake instead of $3x^3y=/=3x^2y$