I need to calculate the fourier transform of this $ t \cdot sin(t-3)+2 \cdot t \cdot cos(3t) \cdot rect(6t) $
is the following valid, based on the first fourier identity i've read in a book, and using linearity
$F\{ (-jt)x(t)\} \quad =\quad \frac { dX(ω) }{ dω } \quad \Longleftrightarrow \\ -j\quad F\{ tx(t)\} =\quad \quad \frac { dX(ω) }{ dω } \quad \Longleftrightarrow \\ F\{ tx(t)\} \quad =\quad j\frac { dX(ω) }{ dω } $
so i can make the problem easier? If not, what is mistaken, and how can i solve this problem? convolution?
Yes, you can use these properties. For the first part $t\sin(t-3)$ just use
$$\sin(t-3)\Longleftrightarrow -je^{-3j\omega}\pi[\delta(\omega-1)-\delta(\omega+1)]=-j\pi[e^{-3j}\delta(\omega-1)-e^{3j}\delta(\omega+1)]$$
which after multiplication by $j$ and differentiation gives
$$t\sin(t-3)\Longleftrightarrow \pi[e^{-3j}\delta^{\prime}(\omega-1)-e^{3j}\delta^{\prime}(\omega+1)]$$
where $\delta^{\prime}(\omega)$ is the derivative of the distribution $\delta(\omega)$. In a similar way you can compute the Fourier transform of $2t\cos(3t)$, which you can then convolve with the Fourier transform of the rectangular function. Just note the following relation for convolution with the derivative of the Dirac impulse:
$$\delta^{\prime}(\omega-\omega_0)*X(\omega)=X^{\prime}(\omega-\omega_0)$$