What you have to understand is that $y$ is a function. You can think of it as $y(x)$. Thus, when you have $xy^2$, $y^2$ is actually a function, which is $[y(x)]^2$. The derivative of $xy^2$ would be:
$$(1)(y^2) + (x)(2)(y)(y'(x))$$
$$y^2 + 2xyy'(x)$$
It is not a constant. Sure, if you have something like $2y$, then the derivative is $2y'(x)$. Notice how if it was $2x$, then its just $2$. But since $y$ is a function , you must treat is as such.
Usually, $y'(x)$, is just abbreviated as $y'$.
Now that you know that, let's differentiate $e^{xy^2}$.
Thats $e^{xy^2}$ * the derivative of ${xy^2}$, which we calculated above. Thus, we have:
$$e^{xy^2} * [y^2 + 2xyy']$$
Differentiating $x - y$ is easy. It's just $1 - y'$.
In the end you have:
$$e^{xy^2} * [y^2 + 2xyy'] = 1 - y'$$
Best Answer
Consider the implicit equation $$F=(y+1)e^y-x=0$$ and compute the partial derivatives $$F'_x=-1$$ $$F'_y= (y+2)e^y$$ Now, from implicit differentiation $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac 1 {(y+2)e^y}$$ So $$y(1.1)\approx y(1)+\frac {1.1-1.0} {(y(1)+2)e^{y(1)}}=0+\frac {0.1} {(0+2)e^{0}}=0.05$$
You could also compute $\frac{dx}{dy}$ and deduce $\frac{dy}{dx}$ from it. This is probably the simplest.