Begin by taking the derivative with respect to #x# of both sides:
#d/dx(9)=d/dx(e^ysin^2x-cos^2y)#
On the left side, we have the derivative of a constant - which is #0#:
#0=d/dx(e^ysin^2x-cos^2y)#
The sum rule says we can break the right side down to:
#d/dx(e^ysin^2x)-d/dx(cos^2y)#
First Term
We are trying to find #d/dx(e^ysin^2x)#. This involves using the product rule, which says:
#d/dx(uv)=u'v+uv'#
In our case, #u=e^y# and #v=sin^2x#. Taking the derivatives of these two, we see:
#u=e^y->u'=e^ydy/dx#
#v=sin^2x=(sinx)^2->v'=2sinxcosx=sin2x#
Substituting into the product rule formula,
#d/dx(e^ysin^2x)=(e^y)'(sin^2x)+(e^y)(sin^2x)'#
#color(white)(X)=e^ysin^2xdy/dx+e^ysin2x#
Second Term
Now we have to find #d/dx(cos^2y)#. Start by rewriting it as #(cosy)^2#. Using the chain rule, we have to take the derivative of the inside term, #cosy# (which is #-sinydy/dx#), and multiply it by the whole derivative (i.e. the derivative of #(cosy)^2#, which is #-2siny#). Doing so gives the result:
#-2siny*-sinydy/dx=2sin^2ydy/dx#
Finally, we can combine our two results into one. Referencing the original equation,
#0=d/dx(e^ysin^2x)-d/dx(cos^2y)#
Substituting what we found,
#0=e^ysin^2xdy/dx+e^ysin2x-2sin^2ydy/dx#
And doing a little algebra to solve for #dy/dx#, we find:
#0=e^ysin^2xdy/dx+e^ysin2x-2sin^2ydy/dx#
#2sin^2ydy/dx-e^ysin^2xdy/dx=e^ysin2x#
#dy/dx(2sin^2y-e^ysin^2x)=e^ysin2x#
#dy/dx=(e^ysin2x)/(2sin^2y-e^ysin^2x)#
