We start from the given equation
#9=e^y sin^2 x - e^x cos^2 y#
Differentiate with respect to x both sides of the equation
#9=e^y sin^2 x - e^x cos^2 y#
#d/dx(9)=d/dx(e^y sin^2 x) - d/dx( e^x cos^2 y)#
Use the product formula #d/dx(uv)=v d/dx(u)+u d/dx(v)# for differentiating the right side of the equation
#0=e^y*d/dx(sin^2 x)+sin^2 x * d/dx(e^y) -[ e^x*d/dx(cos^2 y)+cos^2 y*d/dx(e^x)]#
#0=e^y*2*sin x cos x+sin^2 x * (e^y)y' -[ e^x*2(cos^(2-1) y)(-sin y)y'+cos^2 y*(e^x)*1]#
Simplify
#0=2 e^y sin x cos x+e^y sin^2 x * y'+2e^x sin y cos y y'-e^x cos^2 y#
Solve now for #y'#
Start transposing the terms with #y'# to one side then factor out #y'#
#e^y sin^2 x * y'+2e^x sin y cos y y'=e^x cos^2 y-2 e^y sin x cos x#
#(e^y sin^2 x +2e^x sin y cos y) y'=e^x cos^2 y-2 e^y sin x cos x#
Divide both sides of the equation by the coefficient of #y'#
#((e^y sin^2 x +2e^x sin y cos y) y')/(e^y sin^2 x +2e^x sin y cos y)=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#
#((cancel(e^y sin^2 x +2e^x sin y cos y)) y')/cancel(e^y sin^2 x +2e^x sin y cos y)=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#
# y'=(e^x cos^2 y-2 e^y sin x cos x)/(e^y sin^2 x +2e^x sin y cos y)#
God bless....I hope the explanation is useful.
