Controllability and Observability of a Circuit through State Equation Inspection
Controllability is not just being able to affect the state, it is actually to be able to guide the state from any initial point \$x_0\$ back to the origin \$ x(t_1) = 0\$ in a finite amount of time. And observability is, more or less, being able to deduce the state from previous measurements of the output.
For your system,
$$\dot x_1 = -\frac{1}{L}x_1+u_1 \\ \dot x_2 = -\frac{1}{C}x_2+u_1
$$
with \$u = 0\$, any state would go back to the origin in an infinite amount of time (it would exponentially decay and go to zero but never reach it in a finite amount of time). If you think of the initial point \$x_1 = 1, x_2 = -1\$, you could definitely use the control \$u\$ to drive one of them to zero in a finite time (say \$x_1\$), but then, to also get the other one to zero you would have to use some nonzero \$u\$, which would get \$x_1\$ to stop being \$0\$, and waiting for \$x_2\$ to decay to zero would take infinite time. That would mean that the system is not controllable.
Now, regarding observability, for
$$ y_1 = \frac{1}{L}x_1+\frac{1}{C}x_2+u_1 $$
we can ignore \$u_1\$ (because we will be the ones determining it, so we can just subtract it from \$y_1\$) and look at
$$ y_1 = \frac{1}{L}x_1+\frac{1}{C}x_2, $$
and its derivatives
$$ \dot{y_1} = \frac{1}{L}\dot{x_1}+\frac{1}{C}\dot{x_2} = -\frac{1}{L^2}x_1-\frac{1}{C^2}x_2, \\ \ddot{y_1} = -\frac{1}{L^3}x_1-\frac{1}{C^3}x_2, \\ \vdots $$
since \$C=L\$, we have
$$ y_1 = \frac{1}{L}x_1+\frac{1}{L}x_2 = \frac{1}{L}(x_1+x_2), \\ \dot{y_1} = -\frac{1}{L^2}x_1-\frac{1}{L^2}x_2 = \frac{1}{L^2}(x_1+x_2), \\ \ddot{y_1} = -\frac{1}{L^3}x_1-\frac{1}{L^3}x_2 = \frac{1}{L^3}(x_1+x_2), \\ \vdots $$
meaning that the derivatives of the output will add no information to what we know about the state and from some \$y(t) = c/L\$ we can never know if \$x_1=0,x_2=c\$, or \$x_1=c,x_2=0\$, or \$x_1=-c,x_2=2c\$. Hence we are not able to deduce the state \$x\$ from the output \$y\$, and the system is not observable.