Are there n-th roots of differential operators? In analogy to a Dirac operator, it seems to me that formally, the equation
$$\frac{\partial^n}{\partial x^n}f(x,y)=D_yf(x,y)$$
is solved by 
$$f(x,y)=\exp{(x \sqrt[n]{D_y})}\ g(y).$$
Is there a theory surronding the $\sqrt[n]{D_y}$-idea?
 A: Since you mentioned the Dirac operator, which works not by analysis but by extending the scalars in a noncommutative way; consider
$$\left( \begin{array}{ccc} 0 & 0 & D_y  \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)^3 $$
and generalize.
A: The short answer is yes, absolutely, and the theory of such operators is part of microlocal analysis. The basic ingredient is that differential operators can be written as integral operators (in an appropriate generalized sense) via the Fourier transform. E.g.
$$ \frac{d}{dx} f(x) = \frac{d}{dx} \int e^{2\pi i kx} \hat{f}(k) dk
 = \int e^{2 \pi i k x} (2\pi ik) \hat{f}(k) dk. $$
Since $\hat{f}(k) = \int e^{-2\pi i k y} f(y) dy$ (forgive me if I forgot a $2\pi$ somewhere), we have
$$ \frac{d}{dx} = \int \int (2\pi i k) e^{2 \pi i k(x-y)} dy dk. $$
The right hand side has to be interpreted in a certain distributional sense, but if we are careful such formulae are correct and rigorous. Let's consider your example,
$$ \frac{\partial^n}{\partial x^n} f(x,y) = D_y f(x,y) $$
and let's assume that $D_y$ is an ordinary polynomial differential operator in $y$ with constant coefficients. Since $D_y$ is a polynomial differential operator with constant coefficients, then
$ \widehat{D_y g}(k) = P(k) \hat{g}(k)$ for some polynomial $P$. This suggests that whatever $\sqrt[n]{D_y}$ might be, it should satisfy
$$ \widehat{\sqrt[n]{D_y} g}(k) = \sqrt[n]{P(k)} \hat{g}(k). $$
But using the Fourier transform, we can take this as the definition of $\sqrt[n]{D_y}$:
$$ \sqrt[n]{D_y} g(y) := \int e^{2\pi i ky} \sqrt[n]{P(k)} \hat{g}(k) dk
= \int \int e^{2\pi i k(y-y')} \sqrt[n]{P(k)} g(y') dy' dk. $$
This leads to
$$ \exp(x \sqrt[n]{D_y}) g(y) 
= \int \int e^{2\pi i k(y-y')} \exp(x\sqrt[n]{P(k)}) \hat{g}(k) dy' dk. $$
As long as $P(k)$ and $g(y)$ are nice enough that this expression makes sense (and converges in an appropriate sense), this will solve the given PDE.
