A curve C has equation 2^x + y^2 = 2xy. How do I find dy/dx for the curve C?

This question requires an understanding of implicit differentiation, a specific form of the chain rule for derivatives.
(Aside/reminder) - Implicit function vs Explicit function: Explicit functions: y may be written solely as a function of x. Of the form: y = f(x)e.g. y = 2x + 3Implicit functions:y cannot be solely written as a function of x.Of the form: F(x, y) = 0e.g. e^y + 2xy = 0
Question:1) We differentiate each component of the function with respect to the variables present, using the chain rule, product rule, and other known algebraic rules where necessary.2) For the first component, let z = 2^x . Take the natural logarithm of both sides. We know form logarithmic rules ln(2^x) = xln(2), hence ln(z) = xln(2). Differentiating both sides implicitly using the chain rule we get, dz/dx(1/z) = ln(2). Hence dz/dx = zln(2) = 2^xln(2). This is the derivative of the first component.3) For the second component, let u = y². We know from the chain-rule, du/dx = dy/dx * du/dy. Hence, du/dx = dy/dx * (2y). This is the derivative of the second component.4) For the third component, let w = 2xy. We use the product rule: dw/dx = 2x*(dy/dx * 1) + 2*(y) = dy/dx (2x) + 2y. This is the derivative of the third component.5) Overall the equation differentiates to: 2^xln(2) + dy/dx(2y) = dy/dx*(2x) +2y. This can be rearranged to form dy/dx = [2y - 2^xln(2)]/[2(y-x)].