I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

$3{y}^{2}=5{x}^{3}$

I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:

$\frac{d}{dx}(3{y}^{2})=\frac{d}{dx}(5{x}^{3})$

$6y\frac{d}{dx}(y)=15{x}^{2}\frac{d}{dx}(x)$

$6y\frac{dy}{dx}=15{x}^{2}\frac{dx}{dx}$

$6y\frac{dy}{dx}=15{x}^{2}$

$\frac{dy}{dx}=\frac{15{x}^{2}}{6y}$

Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.