The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted [tex]C[/tex], which we can parameterize by the vector-valued function,
[tex]\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)[/tex]
for [tex]0\le t\le1[/tex], which has differential
[tex]\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt[/tex]
Then with [tex]x(t)=y(t)=z(t)=1-t[/tex], we have
[tex]\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r[/tex]
[tex]=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt[/tex]
[tex]\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt[/tex]
Complete the square in the quadratic term of the integrand: [tex]t^2-2t+2=(t-1)^2+1=(1-t)^2+1[/tex], then in the integral we substitute [tex]u=1-t[/tex]:
[tex]\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt[/tex]
[tex]\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du[/tex]
Make another substitution of [tex]v=u^2[/tex]:
[tex]\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv[/tex]
Integrate by parts, taking
[tex]r=v+1\implies\mathrm dr=\mathrm dv[/tex]
[tex]\mathrm ds=e^v\,\mathrm dv\implies s=e^v[/tex]
[tex]\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv[/tex]
[tex]\displaystyle=-(2e-1)+(e-1)=-e[/tex]
So, we have by the fundamental theorem of calculus that
[tex]\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)[/tex]
[tex]\implies-e=f(1,1,1)-2[/tex]
[tex]\implies f(1,1,1)=2-e[/tex]
