Let's prove that your function is log-convex. The second derivative of the logarithm of your function is (easy to check)
$$\frac{ 2 x \ln (K x)-x^2+1}{(1-x)^3 x}.$$
The denominator is always positive, therefore no problem there. Let's study the positivity of the numerator:
$$2 x \log (K x)-x^2+1 = 2x\ln K + 1-x^2+2x\ln x.$$
The first term is nonnegative for $K\ge 1$, so it's ok, we omit it. So we have $$1-x^2+2x\ln x.$$The value of this function in $x=1$ is zero. It's derivative on $x\in (0,1)$ is $$ 2-2x+2\ln x = 2(1-x+\ln x).$$ To prove that this expression is negative is an easy exercise, therefore we conclude that $1-x^2+2x\ln x+2x\ln K$ is always nonegative on $[0,1]$ - in fact, it's zero in $x=1$ for $K=1$ and strictly greater than zero everywhere if $K>1$. If $K>1$, then $$\frac{ 2 x \ln (K x)-x^2+1}{(1-x)^3 x} $$is strictly positive on $[0,1]$. If $K=1$ then we need to study the following limit:$$\lim_{x\to 1}\frac{ 2 x \ln ( x)-x^2+1}{(1-x)^3 x}. $$ It's possible to show that in is, in fact, $+\infty$ (that what I called an exercise), thus for $K\ge 1 $ $$\frac{ 2 x \ln (K x)-x^2+1}{(1-x)^3 x}>0,\quad 0< x<1. $$ Hence, the function $(Kx)^{x/(1-x)}$ is strictly log-convex, and, in particular, strictly convex.
As to your problem with that second derivative, there's no error there. The logarithm can be indeed negative, but not "sufficiently negative" to counter all positive terms that you had before.
