It's not a particularly challenging derivative, but I would like to know whether or not my approach is correct.
I assumed, prior to the process of differentiation that $y$ is a continuous function of $x$, and hence I applied the chain rule in the instances where $y$ occurred, and also therefore differentiated every term with respect to $x$
So, let's begin:
Term $1$: $\displaystyle\frac{d}{dx}\left(\frac{x}{1-x}\right)=(1-x)^{-2}$
Term $2$: At this point, I did the following: I let $u$ be a continuous function of $x$ such that $u=y^2$, and hence I applied the chain rule, which states that $\displaystyle \frac{du}{dx}=\frac{du}{dy}\times\frac{dy}{dx}$, hence $\displaystyle \frac{d}{dx}(y^2)=2y\frac{dy}{dx}$
Term $3$: $\displaystyle \frac{d}{dx}(3x^3)=9x^2$
Term $4$: Again, the chain rule is applied to attain $\displaystyle \frac{d}{dx}(5y)=5\frac{dy}{dx}$
And from here I solved for $\displaystyle \frac{dy}{dx}$, which is:
$\displaystyle (1-x)^{-2}+9x^2=5\frac{dy}{dx}-2y\frac{dy}{dx}$
$\displaystyle\frac{dy}{dx}=\frac{9x^2}{(5-2y)(1-x)^2}$
So here are the questions; firstly, when differentiating implicit functions, will you always differentiate with respect to only one variable? Secondly, for the chain rule, is it a good idea to substitute, for example; $u=y^2$ or $n=5y$ and then calculate $\displaystyle \frac{du}{dx}$ or $\displaystyle \frac{dn}{dx}$, each time I come across a variable that is not $x$ (in this example)? And lastly, will this approach work for the differentiation of most, if not every, implicit function?
Any responses are appreciated.
Your method is correct, but they are some calculus mistakes at the end :
The same process but on another form :
$$\frac{x}{1-x}-y^2+3x^3=5y \tag 1$$ $$\left(\frac{1}{1-x}+\frac{x}{(1-x)^2}\right)dx -2y\:dy+9x^2dx=5dy \tag 2$$ $\frac{1}{1-x}+\frac{x}{(1-x)^2} = \frac{1}{(1-x)^2}$ $$ (2y+5)dy=\left(\frac{1}{(1-x)^2}+9x^2\right)dx$$ $$ (2y+5)dy=\frac{1+9x^2(1-x)^2}{(1-x^2)}dx$$ $$y'=\frac{dy}{dx}=\frac{1+9x^2(1-x)^2}{(1-x^2)(2y+5)}$$