Step 1: Recall the differentiation formulas

            LINK: https://www.pioneermathematics.com/differentiation-formulae-formula.html

Step 2: Apply the differentiation to the given equation

          EXAMPLE:  log(x + y) - 2xy =0

                              log(x + y) = 2xy = 0

                             \frac{dy}{dx} log(x+y) = \frac{dy}{dx} 2xy

Step 3: Remember the The Product Rule and Differentiate with respect to x

          NOTE:   \frac{dy}{dx} (uv) = u v' + v u' (since u' = \frac{dy}{dx} (u)

                    EXAMPLE:   \frac{dy}{dx} log(x+y) = \frac{dy}{dx} 2xy

                                       \frac{1}{x+y}. \frac{dy}{dx} (x + y) = 2(x \frac{dy}{dx} + v.1)

                                      \frac{1}{x+y}.(1+\frac{dy}{dx}) = \left(2x\frac{dy}{dx}+2y\right)

                                       \frac{1}{x+y} + \frac{dy}{dx} \frac{1}{x+y} - 2x \frac{dy}{dx} ) = 2y  

                                        \frac{dy}{dx} (\frac{1}{x+y} - 2x) = 2y - \frac{1}{x+y}

                                       \frac{dy}{dx} (\frac{(1 - 2x^2 - 2xy )}{x+y} = \frac{2xy + 2y^2 - 1}{x+y}    

                                               \frac{dy}{dx} = \frac{2xy + 2y^2 - 1}{1 - 2x^2 - 2xy}

                       y'(0) = \frac{2 - 1}{1} (since put  x =0 and y =1)

                                      y'(0) = 1