Derivative of $\dfrac{dy^2}{dx}$ I just finish learning the chain rule and am now learning Implicit Differentiation and I am wondering: why is it not possible to take the derivative of $\dfrac{dy^2}{dx}$?  Why do we need to apply the chain rule and find $(\dfrac{dy^2}{dy})\cdot\dfrac{dy}{dx}$?

I understand that $y^2$ is a function and it is equal to $[y(x)]^2$, but basically I am having trouble understanding $\dfrac{dy^2}{dy}$ then $\dfrac{dy}{dx}$.  Why do we find the derivative of ${y^2}$ with respect to $y$, and not $x$?
Note that, if we're working with a function $y=y(x)$, then $y^2$ is just the usual multiplication $y(x)\cdot y(x)=y(x)^2$. This is the same as composing the function
$$
f(x)=x^2
$$
with $y(x)$, which gives $f(y(x))=y(x)^2$.

Now, if we're searching the derivative of $y^2$ with respect to $x$, we must use the Chain Rule and consider $y^2=f(y)$, which gives us
$$
\dfrac{d(y(x)^2)}{dx}=\dfrac{d(f(y(x))}{dx}=\dfrac{df(y)}{dy}\cdot\dfrac{dy}{dx}
$$

And since $\dfrac{df(y)}{dy}=\dfrac{d(y^2)}{dy}=2y$, you get the result. You should notice that since it's an implicit derivative, $\dfrac{dy}{dx}$ will depend on the value of $y(x)$ (that's why it's implicit).

For example, let's suppose that $y(x)$ is given implicitely by the equation $x^2+y(x)^2=1$ and we're trying to find the value of $\dfrac{dy}{dx}$ in a given $x$ point where $y(x)\neq 0$. We derive both sides of the equation, and use the argument given above (before we introduced the example) to get
$$
\dfrac{d}{dx}(x^2+y(x)^2)=0
$$
then
$$
\dfrac{d(x^2)}{dx}+\dfrac{d}{dx}(y(x)^2)=0
$$
$$
2x+2y\dfrac{dy}{dx}=0
$$
This is the part we used the forementioned argument. Now we can get the value we want:
$$
\dfrac{dy}{dx}=\dfrac{-2x}{2y}
$$

I hope it is clear enough!
Given a two fractions, $\frac{a}{b}$ and $\frac{b}{c}$, how do you calculate $\frac{a}{c}$?

Let's relabel $y^2$:

$$\require{cancel}\begin{align}
u&amp;=y^2 \\
\frac{d}{dx}y^2&amp;=\frac{du}{dx}=\cancel{\frac{du}{dy}\frac{dy}{dx}}
\end{align}$$

Take the derivative of $u$ with respect to $y$:
$$\frac{du}{dy}=2y$$