Answer
Yes, $f(1)=\frac{4}{3}$
Work Step by Step
See graph. It seems that the discontinuity at $x=1$ is removable. To find a candidate of $f(1)$, evaluate the limit for $x\to1, x\ne1$:
$\lim_{x\to1}\frac{x-1}{x-\sqrt[4] x}=\lim_{x\to1}\frac{(x-1)(x+\sqrt[4] x)}{(x-\sqrt[4] x)(x+\sqrt[4] x)}=\lim_{x\to1}\frac{(x-1)(x+\sqrt[4] x)}{(x^2-\sqrt x)}=\lim_{x\to1}\frac{(x-1)(x+\sqrt[4] x)(x^2+\sqrt x)}{(x^2-\sqrt x)(x^2+\sqrt x)}=\lim_{x\to1}\frac{(x-1)(x+\sqrt[4] x)(x^2+\sqrt x)}{(x^4- x)}=\lim_{x\to1}\frac{(x-1)(x+\sqrt[4] x)(x^2+\sqrt x)}{x(x- 1)(x^2+x+1)}=\lim_{x\to1}\frac{(x+\sqrt[4] x)(x^2+\sqrt x)}{x((x^2+x+1)}=\frac{(1+\sqrt[4] 1)(1^2+\sqrt 1)}{1((1^2+1+1)}=\frac{4}{3}$
Thus, if we let $f(1)=\frac{4}{3}$, the function will be continuous at $a=1$.
