How do you determine $\dfrac{dy}{dx}$ given $y={{x}^{2}}+xy$ ?

Hint: There are two ways to solve this problem. One is by making the given function explicit and then explicitly differentiating it and the other method is by implicit differentiation of function and then rearranging it to find the required result. We shall go with the explicit differentiation of the given function. Hence, we will differentiate using product rule and chain rule of differentiation.

Complete step by step answer:
Complete step-by-step solution:
The given equation is, $y={{x}^{2}}+xy$and in order to find $\dfrac{dy}{dx}$, we shall differentiate the entire function with respect to x.
Differentiating it with respect to x, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( xy \right)$
We know that the differentiation of any variable, let us say, $t$, raised to some power $n$is given as:
$\dfrac{d}{dt}\left( {{t}^{n}} \right)=n{{t}^{n-1}}$ , applying this on the first term, we get
$\Rightarrow \dfrac{dy}{dx}=2{{x}^{2-1}}+\dfrac{d}{dx}\left( xy \right)$
Also, we use the product rule of differentiation where more than one variable is present in one term to be differentiated. We first differentiate one variable and keep the other ones unchanged, then we add another term in which we keep the first variable unchanged and differentiate the other variable.
Applying this rule in our equation, we get
$\Rightarrow \dfrac{dy}{dx}=2x+1.y+x.\dfrac{dy}{dx}$
$\Rightarrow \dfrac{dy}{dx}\left( 1-x \right)=2x+y$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x+y}{1-x}$ ……………… equation (1)
But result that we have got is in still in terms of y, thus we substitute value of y from $y={{x}^{2}}+xy$and rearranging, we get
$\begin{align}
  & \Rightarrow y\left( 1-x \right)={{x}^{2}} \\
 & \Rightarrow y=\dfrac{{{x}^{2}}}{1-x} \\
\end{align}$
Now, substituting this value in equation (1), we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x+\left( \dfrac{{{x}^{2}}}{1-x} \right)}{1-x}$
We will take LCM in the numerator to further solve:
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \dfrac{2x\left( 1-x \right)+{{x}^{2}}}{1-x} \right)}{1-x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \dfrac{2x-2{{x}^{2}}+{{x}^{2}}}{1-x} \right)}{1-x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \dfrac{2x-{{x}^{2}}}{1-x} \right)}{1-x} \\
\end{align}$
Rearranging further, we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2x-{{x}^{2}}}{{{\left( 1-x \right)}^{2}}}$
Therefore, $\dfrac{dy}{dx}=\dfrac{2x-{{x}^{2}}}{{{\left( 1-x \right)}^{2}}}$ when$y={{x}^{2}}+xy$.

Note:
Another method of solving the question was by making the given function explicit. This could be done by rearranging the initially given equation and separating the x and y-terms, that is, expressing the y-variable in terms of x-variable only and then differentiating the function using the quotient rule of differentiation.