= Solution
Use first the test function $f(x)=x$. The assumed <martingale problem> says that
$$
Y_t=X_t-X_0-\int_0^t b(X_s)\,ds
$$
is a continuous local martingale. Next use $f(x)=x^2$ to see that
$$
X_t^2-X_0^2-\int_0^t\bigl(2X_sb(X_s)+\sigma(X_s)^2\bigr)\,ds
$$
is a local martingale. On the other hand, <Itô formula> applied to $X=X_0+\int b(X_s)ds+Y$ shows that
$$
X_t^2-X_0^2-\int_0^t2X_sb(X_s)\,ds-[Y]_t
$$
is a local martingale. Their difference is both a continuous local martingale and a finite-variation process, so
$$
[Y]_t=\int_0^t\sigma(X_s)^2\,ds.
$$
Part (b), with $H_s=\sigma(X_s)^2>0$, supplies a Brownian motion $W$ such that
$$
Y_t=\int_0^t\sigma(X_s)\,dW_s.
$$
Therefore
$$
X_t=X_0+\int_0^t b(X_s)\,ds+\int_0^t\sigma(X_s)\,dW_s,
$$
so $X$ is a <weak solution of a stochastic differential equation> to the stated equation.
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