From $f(0) = 0$, we conclude that $\color{blue}{a = 0}$. From the fact that $f$ has two local minima and one local maximum, we conclude that $\color{blue}{e > 0}$. Taking the 1st derivative of $f$, we obtain

$$f ' (x) = b + 2 c x + 3 d x^2 + 4 e x^3$$

From $f ' (0) = 0$, we conclude that $\color{blue}{b = 0}$. From $f ' (1) = 0$ and $f (1) = \frac 12$, we obtain a system of $2$ linear equations in unknowns $c$, $d$ and $e$

$$\begin{array}{rl} 2 c + 3 d + 4 e &= 0\\ c + d + e &= \frac 12\end{array}$$

Let us introduce parameter $t > 0$. Let $\color{blue}{e = t}$. Hence, $\color{blue}{c = t + \frac 32}$ and $\color{blue}{d = -1 - 2 t}$. Thus,

$$f ' (x) = (2 t + 3) x + 3 (-1 - 2 t) x^2 + 4 t x^3 = x \left( 2 t + 3 - 3 (1 + 2 t) x + 4t x^2 \right)$$

After some tedious work, we conclude that the maximum is attained at

$$x_{\max} := \frac{2 t + 3}{4t}$$

Since $f (x_{\max}) = 1$, we eventually obtain the quartic equation

$$(2 t + 3)^3 (2 t - 1) - 256 t^3 = 0$$

Using SymPy to solve the quartic equation above:

>>> from sympy import *
>>> t = Symbol('t')
>>> p = (2*t + 3)**3 * (2*t - 1) - 256*t**3
>>> roots = solve(p,t)

Print the $4$ roots in floating-point:

>>> for r in roots:
        r.evalf()


0.401923788646684 - 0.431895218164327*I
0.401923788646684 + 0.431895218164327*I
11.6136005841302
-0.417448161423609

Print the positive root:

>>> roots[2].simplify()
3*sqrt(3)/2 + 3 + sqrt(72 + 42*sqrt(3))/2

Since only one root is real and positive, we conclude that the value of parameter $t$ is

$$\boxed{ \quad t = \frac{3 \sqrt{3}}{2} + 3 + \frac{1}{2} \sqrt{72 + 42 \sqrt{3}} \approx 11.6136005841302 \quad }$$

Let us verify:

>>> from sympy import *
>>> x = Symbol('x')
>>> t = 3*sqrt(3)/2 + 3 + sqrt(72 + 42*sqrt(3))/2
>>> f = (t + 1.5)*x**2 - (1+2*t)*x**3 + t*x**4

Let us check the derivatives:

>>> diff(f,x).subs(x,0)
0
>>> diff(f,x).subs(x,1)
0
>>> diff(f,x).subs(x,(2*t+3)/(4*t))
-3*(3*sqrt(3) + 7 + sqrt(72 + 42*sqrt(3)))*(3*sqrt(3) + 9 + sqrt(72 + 42*sqrt(3)))**2/(6*sqrt(3) + 12 + 2*sqrt(72 + 42*sqrt(3)))**2 + 4*(3*sqrt(3)/2 + 3 + sqrt(72 + 42*sqrt(3))/2)*(3*sqrt(3) + 9 + sqrt(72 + 42*sqrt(3)))**3/(6*sqrt(3) + 12 + 2*sqrt(72 + 42*sqrt(3)))**3 + 2*(3*sqrt(3)/2 + 4.5 + sqrt(72 + 42*sqrt(3))/2)*(3*sqrt(3) + 9 + sqrt(72 + 42*sqrt(3)))/(6*sqrt(3) + 12 + 2*sqrt(72 + 42*sqrt(3)))

The 3rd one is messy. In floating-point:

>>> diff(f,x).subs(x,(2*t+3)/(4*t)).evalf()
0.e-123

which is zero. So far, so good! Let us check if $f (1) = \frac 12$:

>>> f.subs(x,1)
0.500000000000000

It works! Lastly, quartic function $f$ is

$$\boxed{\quad f (x) = \frac 12 \left( 3 \sqrt{3} + 6 + \sqrt{72 + 42 \sqrt{3}} \right) x^{4} - \left(3 \sqrt{3} + 7 + \sqrt{72 + 42 \sqrt{3}}\right) x^{3} + \quad \\ \qquad\quad + \frac 12 \left( 3 \sqrt{3} + 9 + \sqrt{72 + 42 \sqrt{3}} \right) x^{2} \quad}$$

Plotting the graph of function $f$,

The local maximum is attained at

$$\frac{2t+3}{4t} = \frac{1}{2} \left( \dfrac{3 \sqrt{3} + 9 + \sqrt{72 + 42 \sqrt{3}}}{3 \sqrt{3} + 6 + \sqrt{72 + 42 \sqrt{3}}} \right) \approx 0.564579455317661$$