1. We need to evaluate $f(x)=\int \frac{1}{x^{1/4}(1+x^{1/4})}\,dx$ with the condition $f(0)=-6.$ Then we must find $f(1)$.

2. Let $t=x^{1/4} \implies x=t^4, \quad dx=4t^3\,dt.$ Also, $x^{1/4}=t.$ So the integrand becomes $\frac{1}{t(1+t)}\cdot 4t^3\,dt=\frac{4t^2}{1+t}\,dt.$ Hence, $f(x)=\int \frac{4t^2}{1+t}\,dt.$

3. Divide $t^2$ by $(1+t)$: $\frac{t^2}{t+1}=t-1+\frac{1}{t+1}.$ Therefore, $\frac{4t^2}{1+t}=4\left(t-1+\frac{1}{t+1}\right).$ So, $f(x)=4\int \left(t-1+\frac{1}{t+1}\right)dt.$

4. Integrate termwise: $f(x)=4\left(\frac{t^2}{2}-t+\ln(1+t)\right)+C.$ Substituting back $t=x^{1/4}$, $f(x)=2x^{1/2}-4x^{1/4}+4\ln(1+x^{1/4})+C.$

5. Use the condition $f(0)=-6$. At $x=0$, $x^{1/2}=0, \quad x^{1/4}=0, \quad \ln(1+0)=0.$ Thus, $f(0)=C=-6.$ So, $f(x)=2x^{1/2}-4x^{1/4}+4\ln(1+x^{1/4})-6.$

6. Now compute $f(1)$: $f(1)=2(1)-4(1)+4\ln(1+1)-6.$ That is, $f(1)=2-4+4\ln 2-6=4\ln 2-8.$ Factorizing, $f(1)=4(\ln 2-2).$

7. Compare with the options: Option A is $4(\log_e 2-2),$ which matches exactly.

Therefore, the correct answer is A.