(d/du)S[u,x^2]((sin(t)/t)dt) is a representation of the left hand side of the statement. S represents integral while [lower,upper] represents the limits. The right hand side I interpret as (sin(x^2)-sin(u))/u=sin(x^2)/u-sin(u)/u.
We need to prove S[u,x^2]((sin(t)/t)dt)=S((sin(x^2)/u-sin(u)/u)du).
That is, we need to prove the integrands are equal, given the limits.
In this, x^2 has to be considered a constant. So S((sin(x^2)/u)du)=sin(x^2)ln(u) and S((sin(u)/u)du) is effectively the same as S[?,u]((sin(t)/t)dt). The lower limit has not yet been defined.
Rewriting: S[u,x^2]((sin(t)/t)dt)+S[?,u]((sin(t)/t)dt)=sin(x^2)ln(u).
S[0,x^2]((sin(t)/t)dt)=S[u,x^2]((sin(t)/t)dt)+S[0,u]((sin(t)/t)dt)=sin(x^2)ln(u). The lower limit has been set to 0 arbitrarily. Although sin(t)/t is not defined at t=0, when t is small sin(t)=t so the fraction tends to 1.
Another approach: S[u,x^2]((sin(t)/t)dt)=S[u,x^2]((t-t^3/3!+t^5/5!-t^7/7!+...)/t)dt)=
S[u,x^2](1-t^2/3!+t^4/5!-t^6/7!+...)dt)=
(t-t^3/(3*3!)+t^5/(5*5!)-t^7/(7*7!)+...)[u,x^2]=
x^2-u-(x^6-u^3)/(3*3!)+(x^10-u^5)/(5*5!)-(x^14-u^7)/(7*7!)+...
Differentiating wrt u: -1+u^2/3!-u^4/5!+u^6/7!-... This contains no element of x so it cannot be equal to the expression on the right hand side in the original question. The statement would therefore appear to be FALSE.