Your first approach doesn't work because the original signal is simply not the integral of the signal you sketched below it. Just graphically integrate that signal to see this. Your second approach doesn't work for the very same reason. If you integrate the rectangular signal you don't get a signal that is zero for $t>1$. The value of the integral would remain at its final value and wouldn't change anymore because the original signal doesn't change for $t>1$.
That exercise can be solved in several ways, and you should try them all to get some practice. First, just compute the Fourier integral without trying any smart tricks. The signal is simple enough for this to be a good option, even if just as a reference.
Second, convince yourself that if $X(j\omega)$ is the Fourier transform of $x(t)$, then $jX'(j\omega)$ is the Fourier transform of $tx(t)$, where $X'(j\omega)$ is the derivative of $X(j\omega)$ with respect to $\omega$. Since the signal is defined as $t$ times some other signal (whose Fourier transform is very basic), using the above-mentioned property is quite obvious.
A third approach could be to define a signal $x_1(t)$ which satisfies
$$x(t)=\int_{-\infty}^tx_1(\tau)d\tau$$
Then compute the Fourier transform of $x_1(t)$ and use the integration property of the Fourier transform to derive $X(j\omega)$. This is what you actually tried to do, but your definition of $x_1(t)$ was wrong. As a hint, note that for the correct definition of $x_1(t)$ you need two Dirac delta impulses.
Obviously, if you do things right, all three results should agree.
but I wasn't sure how I should go around nesting properties, so I decided that I can perhaps:
However, checking Wolfram, we have:
