Differentiate both sides of the equation with respect to #x#:

#d/dx( sqrtx + sqrt (yx^2) )= d/dx (x)#

#d/dx( sqrtx) + d/dx( sqrt (yx^2) )= 1#

#1/(2sqrtx) + 1/(2sqrt(yx^2)) d/dx (yx^2) = 1#

#1/(2sqrtx) + 1/(2sqrt(yx^2)) (2xy+x^2dy/dx) = 1#

#1/(2sqrtx) + (2xy)/(2sqrt(yx^2)) +x^2/(2sqrt(yx^2))dy/dx = 1#

Note that both #x > 0# and #y > 0# if the equation is in the real domain, otherwise #sqrt x# and #sqrt(yx^2)# would not be defined,

so #x/sqrt(x^2) = 1# and #y/sqrty = sqrty#

#1/(2sqrtx) +sqrty +x/(2sqrt(y))dy/dx = 1#

#x/(2sqrt(y))dy/dx = 1-1/(2sqrtx) -sqrty#

#dy/dx = (2sqrty)/x(1-1/(2sqrtx) -sqrty)#

#dy/dx = (2sqrty)/x-sqrty/(xsqrtx)-(2y)/x#