= Solution
{c}
For $0\leq x\leq1$ and every real $z$,
$$
|\partial_yf(x,z)|=\frac{2x|z|}{(1+z^2)^2}
\leq\frac{x}{1+z^2}\leq1,
$$
where $2|z|\leq1+z^2$ was used. The <mean value theorem> therefore proves <Lipschitz continuity> in $y$ with constant $1$.
Let $e_n=y(x_n)-y_n$. The <explicit Euler method> and the exact-solution identity from part (i) give
$$
e_{n+1}=e_n+h\big(f(x_n,y(x_n))-f(x_n,y_n)\big)+h\tau_n.
$$
Using the requested <local truncation error> bound $|\tau_n|\leq h/2$, including at $n=0$, yields
$$
|e_{n+1}|\leq(1+h)|e_n|+\frac{h^2}{2},\qquad e_0=0.
$$
Iteration and the finite <geometric series> give
$$
|e_N|\leq\frac{h^2}{2}\sum_{j=0}^{N-1}(1+h)^j
=\frac h2\big((1+h)^N-1\big).
$$
Since $1+h\leq e^h$ and $Nh=1$,
$$
\boxed{|y(1)-y_N|\leq\frac{e-1}{2}h.}
$$
Keeping the stronger bound $|\tau_n|\leq h/4$ from part (i) improves the same proof to $(e-1)h/4$. The requested inequality follows already from the weaker bound.
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