Find the implicit derivative of:
$$x^2+x\arcsin(y)=ye^x$$
So taking the derivative results in the function becoming
$$2x+\frac{y^{'}x}{\sqrt{1^2-y^2}}+\arcsin(y)=y^{'}e^x+ye^x$$ Moving all the terms with $y^{'}$to one side gives us
$$\frac{y^{'}x}{\sqrt{1^2-y^2}}-y^{'}e^x=ye^x-2x-\arcsin(y)$$ then factoring out $y^{'}$get us to $$y^{'}\left(\frac{x}{\sqrt{1^2-y^2}}-e^x\right)=ye^x-2x-\arcsin(y)$$ Solving for $y^{'}$ $$y^{'}=\frac{\left(2x+arcsin(y)-ye^x\right) \sqrt{1-y^2}}{{e^x\sqrt{1-y^2}-x}}$$