Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 78

arctan(xy) =arcsin(x + y)
First, take the derivative of both sides of the equation using implicit differentiation.
d/(dx)[arctan(xy)] = d/dx[arcsin(x + y)]
Take note that the derivative formula of arctangent is

d/dx[arctan(u)]=1/(1+u^2)*(du)/dx
And the derivative formula of arcsine is

d/dx[arcsin(u)] = 1/sqrt(1-u^2)*(du)/dx
Applying these two formulas, the equation becomes
1/(1+(xy)^2)*d/dx(xy) = 1/sqrt(1 - (x+y)^2)*d/dx(x+y)
To take the derivative of xy, apply the product rule.

d/dx (u * v) = u *(dv)/dx + v *(du)/dx
Applying this formula, the equation becomes
1/(1+(xy)^2)*(x*d/dx (y) + y*d/dx(x))= 1/sqrt(1 - (x+y)^2)*(d/dx(x)+d/dx(y))
1/(1+(xy)^2)*(x*(dy)/dx + y*1)= 1/sqrt(1 - (x+y)^2)*(1+(dy)/dx)
Then, isolate (dy)/dx .
x/(1+(xy)^2)*(dy)/dx +y/(1+(xy)^2)=1/sqrt(1 - (x+y)^2) +1/sqrt(1 - (x+y)^2)*(dy)/dx
x/(1+(xy)^2) *(dy)/dx - 1/sqrt(1-(x+y)^2)*(dy)/dx = 1/sqrt(1-(x+y)^2) - y/(1+(xy)^2)
(x/(x+(xy)^2)-1/sqrt(1-(x+y)^2))*(dy)/dx= 1/sqrt(1-(x+y)^2) - y/(1+(xy)^2)
(dy)/dx =(1/sqrt(1-(x+y)^2) - y/(1+(xy)^2))/(x/(1+(xy)^2) - 1/sqrt(1-(x+y)^2))
(dy)/dx =(1/sqrt(1-(x+y)^2) - y/(1+(xy)^2))/(x/(1+(xy)^2) - 1/sqrt(1-(x+y)^2)) * (((1+(xy)^2)sqrt(1-(x+y)^2))/1)/(((1+(xy)^2)sqrt(1-(x+y)^2))/1)
(dy)/dx = ((1+(xy)^2)-ysqrt(1-(x+y)^2))/(xsqrt(1-(x+y)^2)-(1+(xy)^2))
(dy)/dx = (1+(xy)^2 - ysqrt(1-(x+y)^2))/(xsqrt(1-(x+y)^2 )-1-(xy)^2)
Then, plug-in the given point to get the slope of the curve at that point. The given point is (0,0).
(dy)/dx= (1+(0*0)^2 -0*sqrt(1 - (x+y)^2))/(0*sqrt(1-(0+0)^2)-1-(0*0)^2)=(1+0+0)/(0-1-0)=1/(-1)=-1
Take note that the slope of a curve at point (x,y) is equal to the slope of the line tangent to that point. So the slope of the tangent line is
m=(dy)/dx = -1
Then, apply the point-slope form to get the equation of the line.
y - y_1 = m(x - x_1)
Plugging in the values, it becomes
y - 0=-1(x - 0)
y = -1(x)
y=-x

Therefore, the equation of the tangent line is y = -x .