Consider the system as shown below

where, y(t)=x(e^t). The system is

The correct option is: linear and non-causal.

We are given a system with input $x\left(t\right)$ and output $y\left(t\right)$ represented by the block diagram in the image, where the system relationship is defined as:
$$y\left(t\right)=x\left(e^t\right)$$
To determine the properties of the system, we analyze its linearity and causality.

1. Linearity Analysis:
A system is linear if it satisfies the properties of superposition and homogeneity. Let's test these properties by applying a linear combination of two arbitrary inputs, $x_1\left(t\right)$ and $x_2\left(t\right)$.
Let:
$$y_1\left(t\right)=x_1\left(e^t\right)$$
$$y_2\left(t\right)=x_2\left(e^t\right)$$
Now, consider a new input $x_3\left(t\right)$ which is a linear combination of the two inputs with constants $a$ and $b$:
$$x_3\left(t\right)=a{x}_1\left(t\right)+b{x}_2\left(t\right)$$
The response of the system to this combined input is:
$$y_3\left(t\right)=x_3\left(e^t\right)$$
Substituting the expression for $x_3\left(t\right)$:
$$y_3\left(t\right)=a{x}_1\left(e^t\right)+b{x}_2\left(e^t\right)=a{y}_1\left(t\right)+b{y}_2\left(t\right)$$
Since the system response to the weighted sum of inputs is equal to the weighted sum of individual responses, the system is linear.

2. Causality Analysis:
A system is causal if its output at any time $t$ depends only on the values of the input at the current time $t$ and/or past times (i.e., for values of the input at $\tau \le t$).
Let us evaluate the system output at a specific time instance, for example, $t=1$:
$$y\left(1\right)=x\left(e^1\right)\approx x\left(2.718\right)$$
Here, the output at the present time $t=1$ depends on the future value of the input signal at $t\approx 2.718$.
Because the output depends on future values of the input for $t>0$, the system is non-causal.

Conclusion:
Thus, the system is both linear and non-causal.