A collection $\{v_1,\dotsc,v_n\}$ in a vector space $V$ is linearly independent if the equation
$$
\lambda_1v_1+\dotsb+\lambda_nv_n=0
$$
is only solved by $\lambda_1=\dotsb=\lambda_n=0$.
In your case, you have the collection $\{f,g,h\}$, which can be viewed as a collection in the vector space of functions $\Bbb R\to\Bbb R$.
Now, note that
$$
1\cdot f+0\cdot g+0\cdot h=f=0
$$
Do you see why this proves that your collection is linearly dependent?
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).
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.
Best Answer
Hint: Linear dependence means we have $af_1(x)+bf_2(x)+cf_3(x)+df_4 (x)=0$ for every $x$ (with not all coefficients $0$), not just for one particular value of $x$.
Put $x=0, x=\pi, x=\pi /2$ and $x =\pi /3$. (Do you know the values of $\sin x$ and $\cos x$ for these values of $x$?). You will get $4$ equations for $a,b,c,d$. Try to show from these equations that $a=b=c=d=0$.