How do you find \[\dfrac{dy}{dx}\] by implicit differentiation of \[{{x}^{3}}-xy+{{y}^{2}}=4\]?

Hint: This question is from the topic of implicit differentiation. In this question, we will the differentiation formula that is \[\dfrac{d}{dx}\left( u\times v \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\]. In solving this question, we will first differentiate the whole function or whole equation with respect to x. After using some differentiation formulas, we will differentiate both the sides of the equation with respect to x. After that, we will find the value of \[\dfrac{dy}{dx}\].

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find \[\dfrac{dy}{dx}\] by implicit differentiation of \[{{x}^{3}}-xy+{{y}^{2}}=4\]. Or, we can say we have to differentiate the given equation with respect to x.
The differentiation of the equation that we have to find using implicit differentiation is
\[{{x}^{3}}-xy+{{y}^{2}}=4\]
Let us differentiate both sides of the above equation with respect to x. We can write the above equation as
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{3}}-xy+{{y}^{2}} \right)=\dfrac{d}{dx}\left( 4 \right)\]
As we know that the differentiation of any constant with respect to x is zero, so we can write
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{3}}-xy+{{y}^{2}} \right)=0\]
The above equation can also be written as
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{3}} \right)-\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=0\]
Using the formula \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\], we can write the above equation as
\[\Rightarrow 3{{x}^{3-1}}-\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=0\]
\[\Rightarrow 3{{x}^{2}}-\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=0\]
Now, we will differentiate the term \[{{y}^{2}}\] with respect to x. For differentiating \[{{y}^{2}}\], we will differentiate normally and write a term in multiplication as \[\dfrac{dy}{dx}\]. So, we can write
\[\Rightarrow 3{{x}^{2}}-\dfrac{d}{dx}\left( xy \right)+2y\dfrac{dy}{dx}=0\]
Now, using the formula of product rule of differentiation that is \[\dfrac{d}{dx}\left( u\times v \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\], we can write
\[\Rightarrow 3{{x}^{2}}-\left( y\dfrac{dx}{dx}+x\dfrac{dy}{dx} \right)+2y\dfrac{dy}{dx}=0\]
We can write the above equation as
\[\Rightarrow 3{{x}^{2}}-y-x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0\]
The above equation can also be written as
\[\Rightarrow 3{{x}^{2}}-y+\left( 2y-x \right)\dfrac{dy}{dx}=0\]
The above equation can also be written as
\[\Rightarrow \left( 2y-x \right)\dfrac{dy}{dx}=y-3{{x}^{2}}\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( y-3{{x}^{2}} \right)}{\left( 2y-x \right)}\]
Hence, we have found \[\dfrac{dy}{dx}\] of \[{{x}^{3}}-xy+{{y}^{2}}=4\] using implicit differentiation. The value of \[\dfrac{dy}{dx}\] we have found is \[\dfrac{y-3{{x}^{2}}}{2y-x}\].

Note:
For solving this type of question, we should have a better knowledge in the topic of implicit differentiation which belongs to the chapter of calculus. We should remember the following formulas:
\[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\]
\[\dfrac{d}{dx}\left( u\times v \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}\]
Always remember that whenever we have to differentiate the term like \[{{y}^{2}}\] with respect to x, then we will first differentiate the term \[{{y}^{2}}\] with respect to y, then we will write the term \[\dfrac{dy}{dx}\].