implicit differentiation is an application on the chain rule which can make your life much much easier than normal, explicit differentiation.
The meaning of explicit differentiation is that u take the derivative of x in terms of x, like saying "I am hungry, and I wanna eat shawarma"
when you are hungry, this means that you explicitly stated that you are hungry and want shawarma.
However, you could ask your friend (if you have friends ofc) "Are you hungry, there is a real good shawarma chain down two blocks away",
this means that you implicitly stated that you are hungry and want shawarma. Lemme give you an example.
$$x^2 + y^2 = 1$$
mnow you can take the explicit route:
$$y = \pm \sqrt{1 - x^2}$$
and divide it into the 2 equations
$$y = + \sqrt{1 - x^2}$$
and
$$y = - \sqrt{1 - x^2}$$
and find \(\frac{dy}{dx}\) by applying the chain rule.
but lets in future examples you most likely wont have the delicacy of doing that, so lets try another approch:
let's differentiate both sides
$$\frac{d}{dx} \left[x^2 + y^2\right] = \frac{d}{dx}1$$
now you know that \(\frac{d}{dx} x^2 = 2x\), but what is \(\frac{d}{dx} y^2\), how could you take the derivative of y in terms of x...
well, think of it like this: you have \(\frac{d}{dx}\), you can multiply both sides of the fraction by \(dy\):
$$\frac{d}{dx} \frac{dy}{dy}$$
which is basically:
$$\frac{d}{dy} \frac{dy}{dx}$$
so it is basically the derivative of y in terms of y times \(\frac{dy}{dx}\), which is exactly the value we are looking for.
so going back to the example:
$$\frac{d}{dx} \left[x^2 + y^2\right] = \frac{d}{dx}1$$
$$ 2x + 2y \cdot \frac{dy}{dx} = 0$$
and from here it is basic algebra:
$$ 2y \cdot \frac{dy}{dx} = -2x$$
$$\frac{dy}{dx} = \frac{-x}{y}$$