With implicit differentiation for finding Dy/dx?

x²y³ - xy = 10
(d/dx)(x²y³ - xy) = d10/dx
(d/dx)(x²y³) - (d/dx)(xy) = 0
x²(dy³/dx) + y³(dx²/dx) - x(dy/dx) - y(dx/dx) = 0
x²(dy³/dy)(dy/dx) + y³(dx²/dx) - x(dy/dy)(dy/dx) - y(1) = 0
x²(3y²)(dy/dx) + y³(2x) - x(1)(dy/dx) - y = 0
3x²y²(dy/dx) + 2xy³ - x(dy/dx) - y = 0
3x²y²(dy/dx) - x(dy/dx) = y - 2xy³
(3x²y² - x)(dy/dx) = y - 2xy³
x(3xy² - 1)(dy/dx) = y(1 - 2xy²)
dy/dx = y(1 - 2xy²) / [x(3xy² - 1)]

　
Let's do each term separately, using product rule and chain rule:

d/dx (x^2 y^3)
= d/dx (x^2) * y^3 + x^2 * d/dx (y^3)
= (2x) * y^3 + x^2 * d/dy (y^3) * dy/dx ----> this last part is chain rule
= 2xy^3 + x^2 * (3y^2) * dy/dx

d/dx (−xy)
= d/dx (−x) * y + (−x) * d/dx (y)
= −1 * y − x * dy/dx
= −y − x * dy/dx


x^2 y^3 − xy = 10

Now differentiate implicitly:

d/dx (x^2 y^3) + d/dx (−xy) = d/dx (10)
2xy^3 + x^2 * (3y^2) * dy/dx − y − x * dy/dx = 0
x^2 * (3y^2) * dy/dx − x * dy/dx = y − 2xy^3
dy/dx (3x^2 y^2 − x) = y − 2xy^3
dy/dx = (y − 2xy^3) / (3x^2y^2 − x)

Write it as f(x,y)=x^2y^3-xy =10
Now find
df = df/dx dx + df/dy dy=0
Here df/dx is partial derivative wrt x. Similarly df/dy.
You have now
(2x y^3 -y) dx + 3x^2 y^2 -x) dy=0
dy/dx = - (2xy^3 -y)/(3x^2 y^2 -x)

Since function is simple you can also directi derivative
df/dx = 2x y^3 - 3x^2 y^2 dy/dx - y - x dy/dx=0
dy/dx = (2x y^3 -y)/(3x^2 y^2 -x)
dy/dx