I’ve a kind outlined in mymodule/src/Type/myForm.php

namespace DrupalmymoduleForm;  use DrupalCoreFormFormBase; use DrupalCoreFormFormStateInterface;  class myForm extends FormBase {  /**  * Specify Drupal Developer kind ID.  */ public perform getFormId() {   return 'mymodule_myform'; }    /**    * Construct Drupal Developer precise kind.    */   public perform buildForm(array $kind, FormStateInterface $form_state) {      // Type API code right here... 

Initially, this routing labored advantageous and Drupal Developer kind displayed at Drupal Developer right path:

mymodule.single_tab:   path: '/node/{node}/edit/efficiency'   defaults:     _form: 'DrupalmymoduleFormmyForm'     _title: 'Add efficiency' 

Then I wanted so as to add a customized entry test, limiting to a particular content material kind:

mymodule.single_tab:   path: '/node/{node}/edit/efficiency'   defaults:     _title: 'Add efficiency'     _controller: 'DrupalmymoduleControllermymoduleController::content material'   necessities:     _custom_access: 'DrupalmymoduleControllermymoduleController::checkAccess' 

Drupal Development entry test works advantageous. Now I feel I’ve to show my kind in Drupal Developer controller:

namespace DrupalmymoduleController;  use DrupalCoreControllerControllerBase; use DrupalCoreFormFormBuilderInterface; use DrupalCoreAccessAccessResult; use DrupalnodeEntityNode;  /**  * Controller for Add Efficiency single tab.  */ class mymoduleController extends ControllerBase {    /**    * Returns Drupal Developer add single efficiency kind.    */   public perform content material() {     $kind = $this->formBuilder->getForm('DrupalmymoduleFormmyForm');         return [           '#type' => 'markup',           '#markup' => $form         ];   } 

This fails to show Drupal Developer kind, and I get:

Name to a member perform getForm() on null 

I attempted defining my kind builder class inside Drupal Developer controller as an alternative, however identical end result. What am I doing improper?