What is Newton's method when there is a lot of input?

I am trying to estimate 3parameters of mixed weibull distribution. And I think the Nweton method for it.

But, error happend

'Error: exit :There are too many output arguments.

Error: etimation_weibull> accelerated_newton (line 30)

if ~ exit ('n_iter','var')

Error: etimation_weibull (line 21)

[x, x_err, x_h] = accelerated_newton (func, dfunc, x0, conv_tol, n_iter);

I think that the error is related to the part where the @ function is used. Is it not possible if there are multiple inputs?

Also, is there any other method for estimating the three parameters of the mixed Weibull distribution?

%% Estimation 3parameters in weibull distribution
load('weibul.mat')

Error using load
Unable to find file or directory 'traj.mat'.

X = Euclidean_distance;  % input data

%% Define first-order and second-order derivatives of Weibull distribution

a = 0.1;
b = 0.1;
c = 0.1;

func = @(x,a,b,c) 1-exp((x-c)/a).^b;
dfunc = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-(x-c)/a).^b;

x0 = Euclidean_distance(1);
conv_tol = 1e-10;
n_iter = 30;


[x, x_err, x_h] = accelerated_newton(func,dfunc,x0,conv_tol,n_iter); %● error happend


%% newton-methods function
function [x, x_err, x_h] = accelerated_newton(func,dfunc,x0,conv_tol,n_iter)

          
         if ~isa(func, 'function_handle') || ~isa(dfunc,'function_handle')
             error('function must be a handle (add"@" before the function name')
         end

           if ~exit('n_iter', 'var')
               n_iter = 30;
           end

          if ~exist('conv_tol', 'var')
              conv_tol = 1e-15;
          end

          x = x0; 
          x_h = nan(n_iter + 1, 1);
          x_h(1) = x;

          for k=1:n_iter
              f = func(x);
              df = dfunc(x);
              delta = f./df;

              if k==1
                  df_23  = dfunc(x - 2.*delta./3);
                  FF = 0.5 + sqrt(max(0 ,0.75.*df_23 - 0.5));
              else
                  d2f = (4.*df + 2.*df_old - 6.*(f - f_old)./(x - x_old))./(x-x_old);
                  A2_hat = -.5.*d2f./df;
                  FF = 0.5 + sqrt(max(0, 0.25 - A2_hat.*delta));
              end
              x_old = x;
              x = x_old - delta./FF;
              x_h(k+1) = x;
              x_err = abs(x_old - x);
              if ((x_err < conv_tol) && (abs(f) - conv_tol))
                  x_h(k+2:n_iter+1) = [];
                  break;
              end

              df_old = df;
              f_old = f;
              if k == n_iter
                  disp('The function did not converge')
              end
          end
end