Answer:
[tex]\frac{dy}{dx} =\frac{(3-2xy+y^{2})}{(x^{2}-2xy)}[/tex]
Step-by-step explanation:
From the equation its is obvious that y is define as an explicit function of x, hence to solve implicitly, we differentiate each term of the equation with respect to x.
[tex]x^{2} y-xy^{2}=3x\\[/tex]
[tex](\frac{d(x^{2})}{dx}y+ \frac{d(y)}{dx}x^{2})-(\frac{d(x)}{dx}y^{2} + \frac{d(y^{2} )}{dx}x)\\(2xy+ \frac{d(y)}{dx}x^{2})-(y^{2} + \frac{d(y^{2} )}{dx}x)=3\\[/tex]
Now [tex]y^{2}[/tex] is a function of y which itself is a function of x. Thus by general differentiation we arrive at
[tex](2xy+x^{2} \frac{dy}{dx})-(y^{2} +2yx\frac{dy}{dx})=3\\[/tex]
[tex]2xy+x^{2} \frac{dy}{dx}-y^{2} -2yx\frac{dy}{dx}=3\\[/tex]
bring similar terms we arrive at
[tex]x^{2} \frac{dy}{dx} -2xy\frac{dy}{dx}=3-2xy+y^{2} \\(x^{2}-2xy)\frac{dy}{dx} =(3-2xy+y^{2})\\\frac{dy}{dx} =\frac{(3-2xy+y^{2})}{(x^{2}-2xy)}[/tex]
