A lot of time we have seen functions defined as $y=f(x)$. This clearly shows that $y$ is a function of the independent variable $x$. But often functions are defined implicitly. For instance, consider the equation $x^2+y^2=25$. Of course this is the equation of circle centered at the center $(0,0)$ with radius $5$. Also circles are not functions. But if we say $y\geq 0$, then the equation describes the upper half-circle which is a function defined by $y=\sqrt{25-x^2}$. Functions defined by equations like $x^2+y^2=25$ are called implicit functions. In some cases like $x^2+y^2=25$, we can easily write an implicit function explicitly as $y=f(x)$, but in many cases we cannot. For example, $x^3+y^3=6xy$. So, we need to devise a way to differentiate an implicit function without writing it as $y=f(x)$. This can indeed be done by the chain rule. You just assume that $y$ is a function of $x$ and use the chain rule. For example,
\begin{align*}
\frac{d}{dx}y^n&=(y^n)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
&=ny^{n-1}\frac{dy}{dx}.
\end{align*}
Let us take a look at another example.
\begin{align*}
\frac{d}{dx}\cos y&=(\cos y)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
&=-\sin y\frac{dy}{dx}.
\end{align*}
Here come more examples.

Example. If $x^2+y^2=25$, find $\frac{dy}{dx}$.

Solution. Differentiating the equation with respect to $x$, we obtain
$$2x+2y\frac{dy}{dx}=0.$$
Solving the resulting equation for $\frac{dy}{dx}$, we obtain
$$\frac{dy}{dx}=-\frac{x}{y}.$$

Example.

1. Find $y’$ if $x^3+y^3=6xy$.

Solution. Differentiate the equation with respect to $x$. Then we obtain
$$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.$$
Solving the resulting equation for $\frac{dy}{dx}$, we obtain
$$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}.$$

2. Find the tangent to $x^3+y^3=6xy$ at $(3,3)$.

Solution. The equation of tangent is
$$y-3=\left[\frac{dy}{dx}\right]_{(3,3)}(x-3).$$
$$\left[\frac{dy}{dx}\right]_{(3,3)}=\frac{2\cdot 3-(3)^2}{3^2-2\cdot 3}=-1.$$ Therefore, the tangent is given by $y=-x+6$.