I can handle $\frac d{dx}$ , but $\frac {d^2}{dx^2}$ ?

$\dfrac{d}{d(x^2)}x^x$
$=\dfrac{dx}{dx^2} \dfrac{e^{x\ln x}}{dx}$
$=\dfrac 1{\frac{dx^2}{dx}} e^{x\ln x} \dfrac d{dx} (x\ln x)$
$=\dfrac 1{2x} x^x (\ln x + 1)$
Substitute $x=2$
$=\dfrac 1 4 2^2 (\ln 2 + 1)$
$=\ln 2 + 1$
$\approx 1.69$

Let $y = x^{2}$ ; then $x^{x} = y^{y^{1/2}/2}$ . Our derivative then becomes

$\frac{d}{d(x^{2})} x^{x} = \frac{d}{dy} y^{y^{1/2}/2}$

If we observe the logarithmic derivative instead, we have by

Chain rule :

$\frac{d}{dy} \ln y^{y^{1/2}/2} = \frac{\frac{d}{dy} y^{y^{1/2}/2}}{y^{y^{1/2}/2}} = \frac{\frac{d}{d(x^{2})} x^{x}}{x^{x}}$

Product rule :

$\frac{d}{dy} \ln y^{y^{1/2}/2} = \frac{d}{dy} \frac{1}{2} y^{1/2} \ln y = \frac{\ln y + 2}{4y^{1/2}} = \frac{\ln x + 1}{2x}$

Equating the two, we get

$\frac{d}{d(x^{2})} x^{x} = x^{x} \frac{\ln x + 1}{2x}$

Thus,

$\left. \frac{d}{d(x^{2})} x^{x} \right\vert_{x=2} = \left. x^{x} \frac{\ln x + 1}{2x} \right\vert_{x=2} = 2^{2} \frac{\ln 2 + 1}{2(2)} \approx \boxed{1.693}$