Proving functions are linearly independent

If you make no assumptions about $a_i$s and then show they must be zero, you have proved linear independence.

From the evaluation at $x=0$ (case 1) you correctly conclude that $a_3=0$. At this point, I suggest simplifying your expression to $a_1x^3+a_2\sin(x)$.

In your evaluation at $x\in(0,1)$ (case 2) you have a logical error of assuming that a sum of numbers being zero requires each number to be zero. $a+b=0$ does not imply $a$ and $b=0$

There is also an error here in the structure of your proof, because if you had shown that all of the $a_i$s are zero, you would be done already.

You have a similar logical error for the evaluation at $x=1$ (case 3).

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It seems you are thinking of the outcome of the values of $a_i$s as depending on what $x$ is, hence the "cases" that are treated independently. But the $a_i$s are always the same, for all $x$. Your idea of evaluating at particular points is a good one, and they build on each other. The evaluation at $x=0$ shows that $a_3=0$. Evaluation at two other particular values of $x$, say, $x=\frac12$ and $x=1$, would give you two equations in two unknowns for $a_1$ and $a_2$, and you could show the only solution is $a_1=a_2=0$.

Alternatively, you could take the derivative of your expression to get $3a_1x^2+a_2\cos(x)=0$, and then evaluation at $x=0$ will give you $a_2=0$ directly, after which you'll be left with $a_1x^3\equiv 0$ from which it isn't hard to show $a_1=0$ either by evaluation at $x=1$ or taking the derivative $3$ times.

Your case 2 and case 3 are wrong.

For case 2, just because $a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$ does not imply that the $a_i$ are all zero. Actually, for any $x \ne 0$, and any $a_2, a_3$, $a_1$ can always be chosen so that $a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$ by setting $a_{1}=-\dfrac{a_{2}sin(x) + a_{3}cos(x)}{x^{3}}$.

For your case 3, $\sin(1) \ne 1$ and $\cos(1) \ne 0$. If you use $\pi/2$, then this works.

Do you know what the Wronskian is? If so, this proof boils down to calculating a $3 \times 3$ determinant.

For $y_1 = x^3$, $y_2 = \sin x$, and $y_3 = \cos x$ the Wronskian is \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{vmatrix}

which evaluates to a non-zero expression. This implies $y_1 = x^3$, $y_2 = \sin x$, and $y_3 = \cos x$ are linearly independent.