The question I am working on is as follows:
Let $y$ be implicitly defined by $$\sin(x-y) - e^{xy} + 1=0$$ and $y(0) = 0$. Find $y''(0)$.
Any help with finding the implicit function and possibly its second derivative is greatly appreciated because I cannot seem to work it out myself.
If you derivate it once you get:
$$\cos(x-y)(1-y') - e^{xy}(xy'+y) =0$$
and for $x=0$ we get $1-y'(0) = y(0)=0$ so $y'(0)=1$ and if we derivate it second time we get:
$$-\sin(x-y)(1-y')^2 -\cos(x-y)y''-e^{xy}(xy'+y)^2 -e^{xy}(xy''+2y')=0 $$
so for $x=0$ we get:
$$ -y''(0)-y(0)^2-2y'(0)=0\implies y''(0)=-2$$