1. Given equation

We are given, for $x \in \mathbb{R},\ x\neq 0$,

$x\int_1^x y(t)\,dt=(x+1)\int_1^x t\,y(t)\,dt.$

Let

$I(x)=\int_1^x y(t)\,dt, \qquad J(x)=\int_1^x t\,y(t)\,dt.$

Then the relation becomes

$xI(x)=(x+1)J(x).$

We need to find $y(x)$.

2. Differentiate the given relation

Using the Fundamental Theorem of Calculus,

$I'(x)=y(x), \qquad J'(x)=x y(x).$

Differentiate

$xI(x)=(x+1)J(x):$

$I(x)+xI'(x)=J(x)+(x+1)J'(x).$

Substitute $I'(x)=y(x)$ and $J'(x)=xy(x)$:

$I(x)+xy(x)=J(x)+(x+1)xy(x).$

So,

$I(x)-J(x)=x^2 y(x).$

3. Use the original equation to eliminate $J(x)$

From

$xI(x)=(x+1)J(x),$

we get

$J(x)=\frac{x}{x+1}I(x).$

Hence,

$I(x)-J(x)=I(x)-\frac{x}{x+1}I(x)=\frac{1}{x+1}I(x).$

But from Step 2,

$I(x)-J(x)=x^2 y(x).$

Therefore,

$\frac{1}{x+1}I(x)=x^2 y(x),$

or

$I(x)=x^2(x+1)y(x).$

Since $I(x)=\int_1^x y(t)dt$, differentiate again:

$I'(x)=y(x)=\frac{d}{dx}\big(x^2(x+1)y(x)\big).$

Now,

$x^2(x+1)=x^3+x^2,$

so

$\frac{d}{dx}\big((x^3+x^2)y\big)=(3x^2+2x)y+(x^3+x^2)y'.$

Thus,

$y=(3x^2+2x)y+(x^3+x^2)y'.$

Rearrange:

$(x^3+x^2)y'+(3x^2+2x-1)y=0.$

That is,

$x^2(x+1)y'+(3x^2+2x-1)y=0.$

4. Convert to first-order separable form

$\frac{y'}{y}=-\frac{3x^2+2x-1}{x^2(x+1)}.$

Now simplify the fraction by partial fractions:

$\frac{3x^2+2x-1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}.$

So,

$3x^2+2x-1=A x(x+1)+B(x+1)+Cx^2.$

Expand:

$=Ax^2+Ax+Bx+B+Cx^2=(A+C)x^2+(A+B)x+B.$

Compare coefficients:

$A+C=3,\qquad A+B=2,\qquad B=-1.$

From $B=-1$, we get $A=3$, and then $C=0$.

Hence,

$\frac{3x^2+2x-1}{x^2(x+1)}=\frac{3}{x}-\frac{1}{x^2}.$

Therefore,

$\frac{y'}{y}=-\left(\frac{3}{x}-\frac{1}{x^2}\right)=-\frac{3}{x}+\frac{1}{x^2}.$

Integrate:

$\int \frac{y'}{y}\,dx=\int\left(-\frac{3}{x}+\frac{1}{x^2}\right)dx.$

So,

$\ln|y|=-3\ln|x|-\frac{1}{x}+\ln|C|.$

Therefore,

$y=Cx^{-3}e^{-1/x}=\frac{C}{x^3}e^{-1/x}.$

5. Match with the options

This is exactly:

$\boxed{\frac{C}{x^3}e^{-1/x}}$

So the correct option is:

C

6. Verification with stored answer

Stored correct answer: C

Our derived answer: C

They agree.