\(y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}\) means \(y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}\)
Then use implicit differentiation on the original equation:
\(y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}\\ y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\)
Left hand side use product rule and chain rule and then right hand side use quotient rule(Wow I used every of them)
Gradient of curve y at point (x_1, y_1)= \(\dfrac{dy}{dx}|_{x=x_1}\)
You substitute \(y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}\) and then express y' in terms of x.
\(y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})+\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})\\ y'=\dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})}{\arctan(e^{3x})}\)
Then you substitute x = 0 for the last step. Do you want me to do it for you? :D
P.S. : I am an Asian
