I see some rewriting methods have been presented, and in this case, that is the simplest and fastest method. But it can also be solved as a fraction using the quotient rule, so for reference, here is a valid method for solving it as a fraction.
Let $f(x) = \frac{\sqrt 2}{t^7}$
Let the numerator and denominator be separate functions, so that $$g(x) = \sqrt2$$ $$h(x) = t^7$$
So $$f(t) = \frac{g(t)}{h(t)}$$
The quotient rules states that $$f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{h^2(t)}$$
Using $$g'(t) = \frac{d}{dt}\sqrt2 = 0$$ $$h'(t) = \frac{d}{dt}t^7 = 7t^6$$
we get, by plugging this into the quotient rule: $$f'(t) = \frac{0\cdot t^7 - \sqrt2\cdot7t^6}{t^{14}}$$
Simplifying this gives us $$\underline{\underline{f'(t) = -\frac{7\sqrt2}{t^8}}}$$
This is also the same as the result you should get by rewriting $$f(t) = \frac{\sqrt2}{t^7} = \sqrt2 \cdot t^{-7}$$ and using the power rule.
hint
Use the Leibnitz formula, which gives the $n^{\text{th}} $ derivative of a product :
$$(f.g)^{(n)} = \sum_{k=0}^n \binom {n}{k} f^{(k)}g^{(n-k)}$$
with
$$f(x)=x+2, \;\; f''=f^{(3)}=...=0$$
and
$$g(x)=(1-x)^{-\frac 13}$$
Best Answer
This is a typical case for using the logarithmic derivative. Denoting this function as $f(x)$, we have: $$\frac{f'(x)}{f(x)}=\frac13\Bigl(2\cdot\frac 1x-\frac1{x+1}\Bigr)=\frac 13\frac{x+2}{x(x+1)}, $$
and therefore $$f'(x)=\frac 13\frac{x+2}{x(x+1)}\biggl(\frac{x^2}{x+1}\biggr)^{\mkern-6mu\frac 13} = \frac{x+2}{3(x+1)\sqrt[3]{x(x+1)}} $$