How do you find $\dfrac{dy}{dx}$ by implicit differentiation of ${{x}^{2}}y+{{y}^{2}}x=-2$ ?

Hint: We need to find the derivative of the given function using implicit differentiation. We start to solve the question by differentiating both sides of the given question. Then, we use product rules in derivatives to find the value of $\dfrac{dy}{dx}$.

Complete step by step solution:
We are given a function and need to find $\dfrac{dy}{dx}$ for the function. We solve the given question with the help of the product rule in derivatives.
In mathematics, the variable y is not always the function of variable x or cannot be written in terms of variable x. In such cases, implicit differentiation helps us to find the derivative of the variable y with respect to variable x without having to solve the equation for y.
According to our question,
$\Rightarrow {{x}^{2}}y+{{y}^{2}}x=-2$
Differentiating both sides of the equation with respect to x, we get,
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}y+{{y}^{2}}x \right)=\dfrac{d}{dx}\left( -2 \right)$
We know that $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}\left( b \right)$
Substituting the same, we get,
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}y \right)+\dfrac{d}{dx}\left( {{y}^{2}}x \right)=\dfrac{d}{dx}\left( -2 \right)$
The product rule in differentiation is given by $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
Applying the product rule to the above equation,
$\Rightarrow \left( {{x}^{2}}\dfrac{d}{dx}\left( y \right)+y\dfrac{d}{dx}\left( {{x}^{2}} \right) \right)+\left( {{y}^{2}}\dfrac{d}{dx}\left( x \right)+x\dfrac{d}{dx}\left( {{y}^{2}} \right) \right)=\dfrac{d}{dx}\left( -2 \right)$
Simplifying the above equation, we get,
$\Rightarrow \left( {{x}^{2}}\dfrac{dy}{dx}+2xy \right)+\left( {{y}^{2}}+2yx\dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -2 \right)$
The derivative of any constant is zero.
Following the same,
$\Rightarrow \left( {{x}^{2}}\dfrac{dy}{dx}+2xy \right)+\left( {{y}^{2}}+2yx\dfrac{dy}{dx} \right)=0$
$\Rightarrow {{x}^{2}}\dfrac{dy}{dx}+2xy+{{y}^{2}}+2yx\dfrac{dy}{dx}=0$
Moving the terms other than $\dfrac{dy}{dx}$ to the other side of the equation,
$\Rightarrow {{x}^{2}}\dfrac{dy}{dx}+2yx\dfrac{dy}{dx}=-2xy-{{y}^{2}}$
Taking $\dfrac{dy}{dx}$ common on the left-hand side of the equation,
$\Rightarrow \left( {{x}^{2}}+2yx \right)\dfrac{dy}{dx}=-2xy-{{y}^{2}}$
We need to isolate $\dfrac{dy}{dx}$ to find its value.
Moving $\left( {{x}^{2}}+2yx \right)$ to the right-hand side of the equation,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-2xy-{{y}^{2}}}{{{x}^{2}}+2yx}$
Taking the minus sign common in the numerator,
$\therefore \dfrac{dy}{dx}=-\dfrac{{{y}^{2}}+2xy}{{{x}^{2}}+2yx}$

Note: We need to know the basic formulae of differentiation to solve this problem easily. Implicit differentiation is useful for finding the derivatives when $y\ne f\left( x \right)$ . Implicit differentiation helps us to compute the derivatives of inverse functions.