[tex]\underbrace{(-3e^x\sin y+3y)}_{M(x,y)}\,\mathrm dx+\underbrace{(3y-3e^x\cos y)}_{N(x,y)}\,\mathrm dy=0[/tex]
You have partial derivatives
[tex]M_y=-3e^x\cos y+3[/tex]
[tex]N_x=-3e^x\cos y[/tex]
which are not equal, so the equation is not exact.
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In case you meant to write
[tex]N(x,y)=3x-3e^x\cos y[/tex]
the equations would be exact, since
[tex]N_x=3-3e^x\cos y[/tex]
and in this case the ODE would be exact. I don't suppose it would hurt to demonstrate how to proceed with solving the ODE...
We're looking for a solution of the form [tex]F(x,y)=C[/tex], which means that upon differentiating we have
[tex]F_x\,\mathrm dx+F_y\,\mathrm dy=0\iff M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0[/tex]
which is to say that we can find [tex]F[/tex] by integrating [tex]M[/tex] and [tex]N[/tex].
[tex]F_x=M\implies F=\displaystyle\int M\,\mathrm dx[/tex]
[tex]F=\displaystyle\int(-3e^x\sin y+3y)\,\mathrm dx[/tex]
[tex]F=-3e^x\sin y+3xy+f(y)[/tex]
Differentiating with respect to [tex]y[/tex], we have
[tex]N=F_y[/tex]
[tex]-3e^x\cos y+3x=3x-3e^x\cos y+f'(y)[/tex]
[tex]0=f'(y)[/tex]
[tex]\implies f(y)=C[/tex]
Thus the general solution (again, assuming you made a typo) would be
[tex]F(x,y)=-3e^x\sin y+3xy+C=C[/tex]
and since both [tex]C[/tex]s can be treated as arbitrary constants,
[tex]F(x,y)=-3e^x\sin y+3xy=C[/tex]
