#### Answer

See below.

#### Work Step by Step

Find the derivative of the function. $$y(x)=\frac{1}{x+4}$$ $$y'(x)=-\frac{1}{(x+4)^2}$$ Substituting these equations into the differential equation yields $$y'=-y^2$$ $$-\frac{1}{(x+4)^2}=-\frac{1}{(x+4)^2}$$ This proves that the equation is a solution to the differential equation and that the solution is valid for all real values.