[tex]x^2\cos y-\sin(y+4x)+\ln(1+x)=0[/tex]
Differentiate both sides with respect to [tex]x[/tex], treating [tex]y[/tex] as a function of [tex]x[/tex]:
[tex]2x\cos y-x^2\sin y\dfrac{\mathrm dy}{\mathrm dx}-\cos(y+4x)\left(\dfrac{\mathrm dy}{\mathrm dx}+4\right)+\dfrac1{1+x}=0[/tex]
[tex]2x\cos y-4\cos(y+4x)-\left(x^2\sin y+\cos(y+4x)\right)\dfrac{\mathrm dy}{\mathrm dx}+\dfrac1{1+x}=0[/tex]
[tex]\left(x^2\sin y+\cos(y+4x)\right)\dfrac{\mathrm dy}{\mathrm dx}=2x\cos y-4\cos(y+4x)+\dfrac1{1+x}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x\cos y-4\cos(y+4x)+\frac1{1+x}}{x^2\sin y+\cos(y+4x)}[/tex]
At the point (0, 0), the derivative is
[tex]\dfrac{\mathrm dy}{\mathrm dx}(0,0)=\dfrac{0-4\cos0+1}{0+\cos0}=-3[/tex]
