If you are on a compact domain $\Omega$in $\mathbb{R}^n$ you need to impose some elliptic boundary conditions or it is difficult to make predictions. Assume for simlicity that $f$ satisfies the Dirichlet boundary conditions, i.e., it vanishes on the boundary of $\Omega$. Then

$$\lambda_3\Vert f\Vert_{L^\infty}=\sup_{x\in\Omega} |D_3f(x)| \leq C\lambda^{3+\frac{n-1}{2}} \Vert f\Vert_{L^2}\leq C'\lambda^{3+\frac{n-1}{2}} \Vert f\Vert_{L^\infty}$$

This gives you

$$\lambda_3\leq C' \lambda^{3+\frac{n-1}{2}}. $$

I personally have serious reasons to believe that the exponent $3+\frac{n-1}{2}$is the best that you can hope for for an arbitrary $D_3$.

If you are on a compact domain $\Omega$in $\mathbb{R}^n$ you need to impose some elliptic boundary conditions or it is difficult to make predictions. Assume for simlicity that $f$ satisfies the Dirichlet boundary conditions, i.e., it vanishes on the boundary of $\Omega$. Then

$$\lambda_3\Vert f\Vert_{L^\infty}=\sup_{x\in\Omega} |D_3f(x)| \leq C\lambda^{\frac{3}{2}} \Vert f\Vert_{L^2}\leq C'\lambda^{\frac{3}{2}} \Vert f\Vert_{L^\infty}$$

This gives you

$$\lambda_3\leq C' \lambda^{\frac{3}{2}}. $$

Edit. In my earlier post I misquoted the result I used. I have fixed the errors. For details see my source X. Bin: Derivatives of the spectral function and Sobolev norms of eigenfunctions on a closed Riemann manifold, Annal. Glob. Annalysis, vol. 26(2004), 231-252, Theorem 1.2.