I am given the following IVP that I am to solve numerically using the Taylor method of order 2: \begin{equation} \left\{ \begin{array}{cc} y' = \underbrace{ye^x + 1}_{f(x,y(x))}, & \; 0\leq x \leq 1 \\ y(0) = 0.5 & \end{array}\right. \end{equation} If I am not wrong, using the Taylor method of order 2 we get: \begin{align*} y_{k+1} &= y_k + hf(x_k,y_k) + \frac{h^2}{2}\left(\frac{\partial f}{\partial x}|_{x=x_k, y=y_k} + \frac{\partial f}{\partial y}|_{x=x_k, y=y_k} f(x_k,y_k) \right) \\ y_{k+1} &= y_k + h(y_ke^{x_k} +1) + \frac{h^2}{2}\left(e^{x_k}(y_k + y_ke^{x_k} + 1) \right) \end{align*}
where for $h = 0.5$, we have $x_0 = 0, x_1 = 0.5, x_2 = 1$ and therefore: \begin{align} y_1 &= 1.5625 \\ y_2 &= 4.2184 \end{align} What I need help with: I am then asked if we were to use another step size of $h = 0.125$ (which is a quarter of the initial $h$), by what factor would we expect (approximately) the error rate to decrease?
My guess is that $E_{h = 0.125} \approx \frac{E_{h = 0.5}}{16}$
Your second derivative expression seems to be wrong at the end. $$ f_x+f_yf=e^xy+e^x(e^xy+1)=e^x((1+e^x)y+1) $$ And yes, it is a second order method like modified or improved Euler, so a factor of $1/16$ when the step size is divided by 4 is to be expected.