Find a process $f=f(t,W_t)$ such that process:
$$X_t=\exp(W_t^2-2tW_t^2)+\int_0^tf(s,W_s)ds$$
is a martingale. Justify the fact that $X_t$ is martingale.
I think I should find a process such that in $dX_t$ there is no drift part ($dt$).
$$dX_t=d(\exp(W_t^2-2tW_t^2))+f(t,W_t)dt$$
$$d(\exp(W_t^2-2tW_t^2))=\exp(W_t^2-2tW_t^2)\Big(-2W_t^2dt+(2W_t-4tW_t)dW_t+(1-2t)dt\Big)$$
Hence $f$ should be equal to
$$f(t,W_t)=\exp(W_t^2-2tW_t^2)\Big(2W_t^2+2t-1\Big)$$
I can justify the above saying that Ito integral (i.e. $\int_0^t \exp(W_s^2-2sW_s^2)(2W_s-4sW_s)dW_s$) is a martingale.
Is this all ok or am I dead wrong?
The principle is clear, one considers $Y_t=\exp((1-2t)W_t^2)$, then Itô's formula yields $$\mathrm dY_t=a(t,W_t)\mathrm dW_t+b(t,W_t)\mathrm dt,$$ for some functions $a$ and $b$, hence $$\mathrm dX_t=\mathrm dY_t+f(t,W_t)\mathrm dt=a(t,W_t)\mathrm dW_t+(b(t,W_t)+f(t,W_t))\mathrm dt,$$ thus, $X$ is a martingale if the drift term is zero, that is, for $$f=-b.$$ Now, $Y_t=g(t,W_t)$ for the regular function $$g:(t,w)\mapsto\exp((1-2t)w^2),$$ hence $\mathrm dY_t=\partial_1g(t,W_t)\mathrm dt+\partial_2g(t,W_t)\mathrm dW_t+\frac12\partial^2_{22}g(t,W_t)\mathrm dt$, that is, $$b=\partial_1g+\tfrac12\partial^2_{22}g.$$ Finally, $$\partial_1g(t,w)=-2w^2g(t,w),$$ and $\partial_2g(t,w)=2(1-2t)wg(t,w)$ hence $$\partial^2_{22}g(t,w)=2(1-2t)g(t,w)+4(1-2t)^2w^2g(t,w),$$ which, unless I made some mistake, does not quite lead to your formula for $f$.