I was studying Thomas's Calculus book and attempted a question using implicit differentiation.
$$x^3=\frac{2x-y}{x+3y}$$
I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ \frac{dy}{dx}=\frac{7y-3x^2(x+3y)^2}{7x} \hspace{1cm}(1)$$
The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ \frac{dy}{dx}=\frac{2-4x^3-9x^2y}{3x^3+1}\hspace{1cm} (2)$$
I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$\frac{dy}{dx}=\frac{7y-3x^4-18x^3y-27x^2y^2}{7x}\hspace{1cm} (1)$$
Replacing $x^3$ in the denominator of $(2)$ by $\frac{2x-y}{x+3y}$ and some simplification gives$$ \begin{align}\frac{dy}{dx}&=\frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\\&=\frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}\end{align}\hspace{1cm} (2)$$
The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$
Therefore $(1)=(2)$. Firstly, I am still not entirely sure $\textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $\textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.
Solving your equation for $y$ we get $$y=\frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.